100 High-Scoring MCQs on Oscillation | Class 11 Physics | Unit 7

100 Important MCQs on Oscillations, Unit 7, Class 10 Physics (Unit-Wise Practice)

This post contains 100 carefully selected multiple-choice questions (MCQs) from the Oscillations unit (Unit 7) of Class 10 Physics, designed according to the FBISE syllabus. These questions include a mix of numerical and conceptual problems, making them exam-ready and high-scoring.

Whether you are preparing for annual board exams or competitive tests, this collection covers:

• Simple harmonic motion (SHM) and its characteristics
• Time period, frequency, and amplitude calculations
• Pendulums, spring-mass systems, and oscillating bodies
• Kinetic and potential energy in SHM
• Damped, forced, and free oscillations
• Resonance phenomena and sharpness of resonance
• High-yield formulas, numerical problems, and conceptual questions

Each question comes with a correct answer and detailed explanation, helping students understand concepts clearly while practicing for exams.

Prepare effectively, revise quickly, and boost your score in the Oscillations unit with these 100 MCQs.

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MCQs No. 1

Total distance travelled by the bob of a simple pendulum in one vibration is equal to:
a) Amplitude
b) Square of amplitude
c) Double of amplitude
d) Quadruple of amplitude

Correct Answer is option d. Quadruple of amplitude

Explanation: 

Let amplitude = A. One complete vibration comprise of four amplitude ie.  O to X → X to O → O to X' → X' to O = 4A.

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MCQs No. 2

The period of a simple pendulum is independent of:
a) Length
b) Mass
c) Acceleration due to gravity
d) Amplitude (small)

Correct Answer is option b. Mass

Explanation: 

$T = 2\pi\sqrt{L/g}$

mass has no effect.

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MCQs No. 3

For a simple pendulum, the time period is minimum when:
a) Amplitude is maximum
b) Amplitude is small
c) Amplitude is medium
d) Amplitude is large

Correct Answer is option b. Amplitude is small

Explanation: 

SHM approximation holds only for small amplitude.

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MCQs No. 4

Frequency of the second’s pendulum is_______________.
a. 2Hz,
b. 1Hz,
c. 0.2Hz
d. None of these

The Correct Answer is d. None of these

Explanation:
As f= 1/T =1/2 = 0.5 Hz.

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MCQs No. 5

Simple harmonic motion is a type of:
a. Rotational motion
b. Circular Motion
c. Musical arrangement
d. Oscillatory

The Correct Answer is option d. Oscillatory 

Explanation:

The repeated to and fro motion of a body about mean position is called Oscillatory Motion.

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MCQs No. 6

A body in simple harmonic motion makes n complete oscillations in one second. The angular frequency of this motion is:

a) $\pi \,\text{rad s}^{-1}$

b) ${\displaystyle \frac{1}{\pi }}{\text{rad s}}^{-1}$

c) $2\pi n \,\text{rad s}^{-1}$

d) $\dfrac{2}{2\pi}\,\text{rad s}^{-1}$

The Correct Answer is option c. 2𝜋 𝑛 𝑟𝑎𝑑/𝑠

Explanation:

Angular frequency is defined as the rate of change of angular displacement:

$$\omega = \frac{\Delta \theta}{\Delta t}$$

For one complete oscillation, angular displacement is:

$$\Delta \theta = 2\pi \,\text{rad}$$

If the body makes n oscillations per second, then:

$$\omega = n \times 2\pi$$

$$\boxed{\omega = 2\pi n \,\text{rad s}^{-1}}$$

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MCQ No. 7

If the length of a simple pendulum is quadrupled, its period becomes:
a) Double
b) Four times
c) Half
d) Unchanged

Correct Answer: a. Double

Explanation: $T \propto \sqrt{L}$

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MCQ No. 8

In SHM, the acceleration is:
a) Constant
b) Zero
c) Time dependent
d) Always directed towards mean position

Correct Answer is option d. Always directed towards mean position

Explanation: Restoring acceleration in SHM always toward equilibrium.

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MCQ No. 9

The phase difference between displacement and velocity in SHM is:
a) π/2
b) π
c) 0
d) 2π

Correct Answer is option a. π/2

Explanation: 

Velocity leads displacement by 90°.

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MCQs No.10
A body is moving in a circle with uniform speed. Its motion is:
a. Periodic & simple motion
b. Periodic but not simple harmonic
c. Variable
d. None

The Correct Answer is option b. Periodic but not simple harmonic

Explanation:
Every harmonic motion is a SHM but not every periodic motion is SHM.

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MCQs No.11

Which of the following is the SI base unit of spring constant?

a) $\mathrm{kg\,s^{-2}}$
b) $\mathrm{kg\,m^{-1}}$
c) $\mathrm{kg\,m\,s^{-2}}$
d) $\mathrm{kg\,m^{2}}$

The Correct Answer is option a. kgs−2

Explanation:

Spring constant is defined as:

$$k = \frac{F}{x}$$

Unit of force (Newton):

$$1\,N = \mathrm{kg\,m\,s^{-2}}$$

Now,

$$[k] = \frac{N}{m}$$

​ $$[k] = \frac{\mathrm{kg\,m\,s^{-2}}}{m}$$

​ $$[k] = \mathrm{kg\,s^{-2}}$$
$$\boxed{\text{SI base unit of spring constant} = \mathrm{kg\,s^{-2}}}$$

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MCQs No.12
The SI unit of force constant is identical with that of:
a. Force
b. Pressure
c. Surface tension
d. Loudness

The Correct Answer is option c. Surface tension

Explanation:

Force constant (spring constant) is defined as:

$$k = \frac{F}{x}$$

SI unit of force constant:

$$[k] = \frac{N}{m}$$

Surface tension is defined as force per unit length, so its unit is also:

$$\frac{N}{m}$$

Since both force constant and surface tension have the same SI unit $\text{N m}^{-1}$, their units are identical.

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MCQ No. 13

Maximum kinetic energy in SHM occurs at:
a) Mean position
b) Extreme position
c) Quarter position
d) Half amplitude
Correct Answer: a. Mean position

Explanation: 

KE maximum where PE minimum (x = 0).

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MCQ No. 14

If the time period of a pendulum at place A is T, at B (where g is 4g), the period will be:
a) T/2
b) 2T
c) 4T
d) T
Correct Answer: a. T/2

Explanation: $T \propto 1/\sqrt{g}$.

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MCQ No. 15

In SHM, the total energy is:
a) Constant
b) Variable
c) Zero
d) Infinite
Correct Answer is option a. Constant

Explanation: Total energy remains constant (KE + PE).

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MCQs No.16

When n springs of the same stiffness are connected in parallel, the equivalent spring constant is:

a) $K$
b) $\dfrac{K}{n}$
c) $nK$
d) None of these

The Correct Answer is Keq= n k 

Explanation:

For springs connected in parallel, the equivalent spring constant is the sum of individual spring constants:

$$K_{\text{eq}} = k_1 + k_2 + k_3 + \cdots$$

If all springs have the same stiffness $K$ and their number is n, then:

$$K_{\text{eq}} = K + K + K + \cdots \text{(n times)}$$$$\boxed{K_{\text{eq}} = nK}$$

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MCQs No.17

When springs are connected in series, the equivalent spring constant is:

a) $K_{\text{eq}} = nK$
b) $K_{\text{eq}} = \dfrac{K}{n}$
c) $K_{\text{eq}} = n^{2}K$
d) None of these

The Correct Answer is option c.  $K_{\text{eq}}={\displaystyle \frac{K}{n}}$

Explanation:

For springs connected in series, the reciprocal of the equivalent spring constant is the sum of reciprocals:

$$\frac{1}{K_{\text{eq}}} = \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} + \cdots$$

For two identical springs of constant $K$:

$$\frac{1}{K_{\text{eq}}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$$

​ $$K_{\text{eq}} = \frac{K}{2}$$

Similarly, for n identical springs:

$$\frac{1}{K_{\text{eq}}} = \frac{n}{K}$$

​ $$\boxed{K_{\text{eq}} = \frac{K}{n}}$$

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MCQs No.18

Two identical springs of constant k are connected in Series and then in Parallel. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be?
a. 2:1
b. 1:1
c. 1:2
d. 4:1

The Correct Answer is option c. 1:2

Explanation:

The frequency of a mass–spring system is given by:

$$f = \frac{1}{2\pi}\sqrt{\frac{k_{\text{eq}}}{m}}$$For springs in series:$$\frac{1}{k_s} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$$

​ $$k_s = \frac{k}{2}$$

​ $$f_s = \frac{1}{2\pi}\sqrt{\frac{k/2}{m}}$$For springs in parallel:$$k_p = k + k = 2k$$

$$f_p = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}$$

Ratio of frequencies:

$$\frac{f_s}{f_p} = \sqrt{\frac{k/2}{2k}}$$ $$\frac{f_s}{f_p} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$
$$\boxed{f_s : f_p = 1 : 2}$$

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MCQ No. 19

A particle executes SHM with amplitude 0.20 m and angular frequency 10 rad s⁻¹. Find its maximum velocity.

a) 1 m s⁻¹
b) 2 m s⁻¹
c) 4 m s⁻¹
d) 10 m s⁻¹

The Correct Answer is option b. 2 m s⁻¹

Explanation:

$$v_{\max} = \omega A$$

$$v_{\max} = 10 \times 0.20 = 2 \,\text{m s}^{-1}$$

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MCQ No. 20

If amplitude of SHM doubles, its total energy becomes:
a) Double
b) Four times
c) Half
d) Same

Correct Answer is option b. Four times

Explanation: 

Total energy $E \propto A^2$

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MCQs No. 21

A particle in SHM passes through equilibrium with velocity v. Its maximum velocity is:
a) v
b) 2v
c) v/2
d) √2 v

Correct Answer: a. v
Explanation: At Equilibrium position the velocity is max.

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MCQs No.22

A weight is suspended from an ideal spring oscillates up and down with period T. If the amplitude of the oscillation is doubled, the period will be?

a. $\dfrac{T}{2}$

b. $2T$

c. $\sqrt{2}\mathrm{<br>}$

d. T

The Correct Answer is option d. T

Explanation:

The time period of a spring–mass system is independent of amplitude.

$$T=2\pi \sqrt{\frac{m}{k}}$$

Since amplitude does not appear in the formula, doubling it does not change the period.

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MCQs No.23

A particle moves such that its acceleration a is given by

$$a = -bx$$

where x is the displacement from the equilibrium position and b is a constant. The period of oscillation is:

a) $2\pi\sqrt{b}$

b) $\dfrac{2\pi}{\sqrt{b}}$

c) $\dfrac{2\pi}{b}$

d) $2\sqrt{\dfrac{\pi}{b}}$

Correct Answer is option 

Correct Answer is option b. 

${\displaystyle \frac{2\pi }{\sqrt{b}}}$

    

Explanation :

In simple harmonic motion, acceleration is given by:

$$a = -\omega^2 x$$

Comparing with the given equation:

$$a = -bx$$

We get:

$$\omega^2 = b$$

$$\omega =\sqrt{b}$$

The time period of SHM is:

$$T = \frac{2\pi}{\omega}$$

Substituting $\omega = \sqrt{b}$

$$\boxed{T = \frac{2\pi}{\sqrt{b}}}$$

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MCQ No. 24

In SHM, displacement is zero at:
a) Mean position
b) Extreme position
c) Quarter amplitude
d) Any time

Correct Answer: a. Mean position
Explanation: By definition, equilibrium point or mean position.

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MCQ No. 25

The acceleration of a particle in SHM is:
a) Maximum at equilibrium
b) Zero at extreme position
c) Zero at mean position
d) Maximum at mean position

Correct Answer is option c. Zero at mean position
Explanation: Acceleration = –ω²x; x=0 at mean.

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MCQ No. 26

The maximum velocity in SHM is:
a) ωA
b) ω/2
c) A/ω
d) ω²A

Correct Answer: a. ωA
Explanation: $v_{\max} = \omega A$

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MCQs No.27

If the period of oscillation of mass (m) suspended form a spring is 2s, then the period of mass 4 m will be:
a. 1s
b. 2s
c. 3s
d. 4s

The Correct Answer is option d. 4s

Explanation:

For a mass–spring system, the time period is given by:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

Time period is proportional to the square root of mass:

$$T \propto \sqrt{m}$$

If the mass is increased from $m$ to $4m$:

$$T' = T \sqrt{\frac{4m}{m}} = T \sqrt{4} = 2T$$

Given $T = 2\,\text{s}$:

$$T' = 2 \times 2 = 4\,\text{s}$$

$$\boxed{T' = 4\,\text{s}}$$

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MCQs No.28

If the angular frequency of the oscillating mass-spring system is 2π then the time period is?
a. 1s
b. 2s
c. 3s
d. 4s

The Correct Answer is option a. 1s 

Explanation:

The relation between time period and angular frequency is:

$$\omega = \frac{2\pi}{T}$$

Given:

$$\omega = 2\pi$$

Substituting:

$$2\pi = \frac{2\pi}{T}$$

​ $$T = 1\,\text{s}$$

$$\boxed{T = 1\,\text{s}}$$

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MCQ No. 29

If the period of a simple pendulum is 2 s, its frequency is:
a) 0.5 Hz
b) 2 Hz
c) 4 Hz
d) 1 Hz

Correct Answer is option a. 0.5 Hz

Explanation: 

f = 1/T = 1/2 = 0.5 Hz

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MCQ No. 30

The time period of a mass–spring system is independent of:
a) Mass
b) Spring constant
c) Amplitude
d) Frequency

Correct Answer: c. Amplitude
Explanation: $T = 2\pi\sqrt{m/k}$

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MCQ No. 31

The potential energy of SHM is zero at:
a) Mean position
b) Extreme position
c) Half amplitude
d) None

Correct Answer: a. Mean position

Explanation:

The potential energy of a particle in simple harmonic motion is given by:

$$P.E = \frac{1}{2}kx^{2}$$

At the mean (equilibrium) position, the displacement $x = 0$.

Substituting $x = 0$:

$$P.E = \frac{1}{2}k(0)^2 = 0$$

$$\boxed{\text{Therefore, the potential energy is zero at the mean position.}}$$

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MCQs No.32

When an object is moving in simple harmonic motion, which of the following is at a minimum when the displacement from equilibrium is zero?
a. Magnitude of the velocity
b. Magnitude of the acceleration
c. Kinetic energy
d. Total mechanical energy

The Correct Answer is option b. Magnitude of the acceleration

Explanation:

In simple harmonic motion, the acceleration and -ive displacement are directly related:

$$a = -\omega^{2} x$$

At the equilibrium (mean) position, the displacement $x = 0$.

$$a = -\omega^{2}(0) = 0$$

Thus, the magnitude of acceleration is minimum (zero) at the equilibrium position.

$$\boxed{\text{Hence, the magnitude of acceleration is minimum when displacement is zero.}}$$

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MCQs No.33

The displacement of a particle at time $t$ is given by:

$$x = 10 \sin 4t$$

The particle oscillates with a period of:

a) $\dfrac{\pi}{10}\,\mathrm{s}$ 

b) $\dfrac{\pi}{5}\,\mathrm{s}$ 

c) $\dfrac{\pi}{4}\,\mathrm{s}$ 

d) $\dfrac{\pi}{2}\,\mathrm{s}$ 

The Correct Answer is option d.  ${\displaystyle \frac{\pi }{2}}\mathrm{s}$ 

Explanation:

Compare the given equation with standard SHM:

$$x = x_0 \sin (\omega t)$$

Here, $\omega = 4\,\mathrm{rad/s}$ 

The time period of SHM is:

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2}\,\mathrm{s}$$

$$\boxed{{\displaystyle T=\frac{\pi }{2}\mathrm{s}}}$$

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MCQ No. 34

In a simple pendulum, if length decreases, period:
a) Increases
b) Decreases
c) Remains same
d) Doubles

Correct Answer: b. Decreases
Explanation: $T \propto \sqrt{L}$

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MCQ No. 35

If a pendulum is taken to the moon where g is smaller, its period will:
a) Increase
b) Decrease
c) Stay same
d) Zero

Correct Answer: a. Increase
Explanation: 

$T \propto 1/\sqrt{g}$

Smaller g ⇒ larger T.

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MCQs No. 36

For SHM displacement equation x = A sin ωt, velocity equation is:
a) v = ωA sin ωt
b) v = ωA cos ωt
c) v = A cos ωt
d) v = A tan ωt

Correct Answer is option b. v = ωA cos ωt

Explanation: 

Derivative of sin is cos.

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MCQs No.37

A mass attached to a spring is starting from its mean position. The time at which the mass is at a position half of its amplitude is?
a. T/2
b. T/3
c. T/6
d. T/12

The Correct Answer is option d. T/12 

Explanation:

Since the mass starts from the mean position, the displacement in SHM is:

$$x = A \sin(\omega t)$$

Given that the position is half of the amplitude:

$$x = \frac{A}{2}$$

Substitute:

$$\frac{A}{2} = A \sin(\omega t) \Rightarrow \sin(\omega t) = \frac{1}{2}$$

​ $$\omega t = \frac{\pi}{6}$$

But

$$\omega = \frac{2\pi}{T}$$

So,

$$t = \frac{\pi/6}{2\pi/T} = \frac{T}{12}$$

​ $$\boxed{t = \frac{T}{12}}$$

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MCQs No.38

An object undergoes S.H.M has maximum speed when its displacement from the mean position is:
a. Maximum speed
b. Zero
c. Half of the maximum value
d. 1/3rd of the maximum value

The Correct Answer is option b. Zero 

Explanation:

In simple harmonic motion, the speed of the object is given by:

$$v = \omega \sqrt{A^{2} - x^{2}}$$

The speed is maximum when the displacement from the mean position is zero $(x = 0)$.

At the mean position:

• Kinetic energy is maximum
• Potential energy is minimum
• Speed is maximum

$$\boxed{\text{Therefore, the maximum speed occurs at zero displacement.}}$$

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MCQ No. 39

In SHM, phase difference between velocity and acceleration is:
a) π/2
b) π
c) 0
d) π/4

Correct Answer: b. π

Explanation: 

Velocity leads acceleration by 180°.

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MCQ No. 40

Restoring force in SHM is:
a) Constant
b) Zero
c) Proportional to displacement
d) Proportional to square of x

Correct Answer: c. Proportional to displacement
Explanation: F = –kx.

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MCQ No. 41

The maximum potential energy in SHM occurs at:
a) Mean position
b) Extreme position
c) Half amplitude
d) At t = 0

Correct Answer: b. Extreme position
Explanation: 

The potential energy of a particle in simple harmonic motion is given by:

$$P\mathrm{.}E=\frac{1}{2}k{x}^{}$$

PE max when x = A.

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MCQs No. 42

The phase of SHM at mean position is:
a) 0
b) π/2
c) π
d) 2π

Correct Answer: a. 0
Explanation: By standard definition.

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MCQs No.43

The period of simple pendulum doubles when:
a. Its length is doubled
b. The mass of the bob is doubled
c. Its length is made four times
d. The mass and length of the pendulum is made two times

The Correct Answer is option c. Its length is made four times 

Explanation:

The time period of a simple pendulum is given by:

$$T=2\pi \sqrt{\frac{L}{g}}$$

The period depends only on the length of the pendulum.

If the length is made four times:

$$L' = 4L$$

$$T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times 2\pi \sqrt{\frac{L}{g}} = 2T$$
$$\boxed{{\displaystyle \text{<span class="vlist"><span class="boxpad"><span class="mord"><span class="mord"><span class="mord text"><span class="mord">Thus, the period doubles when the length is made four times.</span></span></span></span></span><span class="pstrut"></span><span class="stretchy fbox"></span></span><span class="vlist-s"></span>}}}$$

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MCQs No.45

A simple pendulum performs simple harmonic motion about  $x=0$ with amplitude $A$ and time period $T$. The speed of the pendulum at $x = \dfrac{A}{2}$ is:

a) $\dfrac{\pi A\sqrt{3}}{T}$

b) $\dfrac{\pi A}{T}$

c) $\dfrac{\pi A\sqrt{3}}{2T}$

d) $\dfrac{3r^{2}A}{T}$

Correct Answer is option a.

 $\dfrac{\pi A\sqrt{3}}{T}$

Explanation:

The speed of a particle executing simple harmonic motion is given by:

$$v = \omega \sqrt{A^{2} - x^{2}}$$

where:

$$\omega = \frac{2\pi}{T}$$

At $x = \dfrac{A}{2}$:

$$v = \frac{2\pi}{T} \sqrt{A^{2} - \left(\frac{A}{2}\right)^{2}}$$$$v = \frac{2\pi}{T} \sqrt{\frac{3A^{2}}{4}}$$$$v = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2}$$$$\boxed{{\displaystyle v=\frac{\pi A\sqrt{3}}{T}}}$$

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MCQs No.46

The period of oscillation of a simple pendulum of constant length at earth poles is T. Its period at equator is:
a. Greater than T
b. Less than T
c. Equal to T
d. Cannot be compared

The Correct Answer is option a. Greater than T 

Explanation:

The time period of a simple pendulum is given by:

$$T=2\pi \sqrt{\frac{L}{g}}$$

As value of g at equator is smaller than at poles. So, the Time period will be Greater than T at poles.

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MCQ No. 47

If frequency of SHM is 50 Hz, period is:
a) 0.02 s
b) 2 s
c) 0.5 s
d) 50 s

Correct Answer: a. 0.02 s

Explanation: 

T = 1/f = 1/50 = 0,02 Hz

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MCQ No. 48

A body in SHM passes through the mean with speed 10 m/s. If amplitude is 0.2 m, ω is:
a) 50 rad/s
b) 25 rad/s
c) 100 rad/s
d) 10 rad/s

Correct Answer: a. 50 rad/s
Explanation: $v = \omega A ⇒ 10=ω(0.2) ⇒ ω=50$

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MCQ No. 49

A mass of 1 kg is attached to a spring of constant 400 N m⁻¹. Find the time period of oscillation.

a) 0.31 s
b) 0.62 s
c) 1.0 s
d) 2.0 s

The Correct Answer is option b. 0.62 s

Explanation:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

​

$$T = 2\pi \sqrt{\frac{1}{400}} = 2\pi \times \frac{1}{20} = 0.628\,\text{s} \approx 0.62\,\text{s}$$

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MCQs No.50

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is:
a. Greater than T
b. Less than T
c. Equal to T
d. Cannot be compared

The Correct Answer is option a. Greater than T

Explanation:

The time period of a simple pendulum is given by:

$$$$T=2πLg

As the value of g decrease inside a mine. So, the Time period will be Greater than T.

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MCQs No.51
The exact length of second pendulum is _________?
a. 100 cm
b. 99 cm
c. 99.2 cm
d. 98 cm

The Correct Answer is option c. 99.2 cm 

Explanation:

A second pendulum is defined as a pendulum whose time period is 2 seconds.

The time period of a simple pendulum is given by:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

For a second pendulum:

$$T = 2\,\text{s}$$

Substituting:

$$2 = 2\pi \sqrt{\frac{L}{g}} \Rightarrow 1 = \pi \sqrt{\frac{L}{g}}$$ $$\sqrt{\frac{L}{g}} = \frac{1}{\pi} \Rightarrow \frac{L}{g} = \frac{1}{\pi^{2}}$$

​ $$L = \frac{g}{\pi^{2}}$$

Taking $g = 9.8\,\text{m/s}^2$:

$$L = \frac{9.8}{\pi^{2}} \approx 0.992\,\text{m} = 99.2\,\text{cm}$$$$\boxed{L = 99.2\,\text{cm}}$$

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MCQ No. 52

The displacement of a particle in SHM is given by

$$x = 0.05 \sin (20t)$$

Find the amplitude of motion.

a) 0.20 m
b) 0.05 m
c) 20 m
d) 1 m

The Correct Answer is option b. 0.05 m

Explanation:
In SHM equation $x = A \sin(\omega t)$ ,
Coefficient of sine = Amplitude

$$A = 0.05\,\text{m}$$

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MCQs No. 53

In SHM, if ω doubles, period:
a) Halves
b) Doubles
c) Quadruples
d) Unchanged

Correct Answer: a. Halves
Explanation: $T = 2\pi/ω$

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MCQs No.54

A simple pendulum is suspended from the roof of a lift. When the lift is moving upward with acceleration $a$ ($a < g$ ), the time period of the pendulum is given by:

$$T = 2\pi \sqrt{\frac{L}{a'}}$$

where the effective acceleration '$a\text{'}$ is equal to:

a)  $g$

b) $g - a$

c)  $g+a$

d) $\sqrt{g^{2} + a^{2}}$

The Correct Answer is option c. g + a

Explanation:

In upward direction net acceleration increases.

When the lift accelerates upward, the effective gravitational acceleration acting on the pendulum increases.

$$a' = g + a$$

Since the time period of a simple pendulum depends on effective gravity, it becomes:

$$T = 2\pi \sqrt{\frac{L}{g + a}}$$ $$\boxed{\text{Thus, } a' = g + a}$$

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MCQs No.55

A simple pendulum suspended from the ceiling of a lift has time period T, when the lift is at rest. When the lift falls freely, the time period is:
a. Infinite
b. Increased
c. Zero
d. Remains the same

The Correct Answer is option a. Infinite

Explanation:

The time period of a simple pendulum is given by:

$$T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$$

When the lift falls freely, it has a downward acceleration equal to g. Hence, the effective acceleration due to gravity inside the lift becomes:

$$g_{\text{eff}} = g - g = 0$$

Substituting $g_{\text{eff}} = 0$  in the formula:

$$T = 2\pi \sqrt{\frac{l}{0}} \rightarrow \infty$$

Therefore, the pendulum does not oscillate, and its time period becomes infinite.

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MCQs No. 56

The time period of a physical pendulum depends on:
a) Mass only
b) Distribution of mass
c) Gravity only
d) Amplitude

Correct Answer: b. Distribution of mass
Explanation: Time period T of physical pendulum depends on moment of inertia.

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MCQ No. 57

In SHM, energy versus time graph is:
a) Straight line
b) Sinusoidal
c) Constant for total energy
d) Parabolic

Correct Answer: c. Constant for total energy
Explanation: Total energy E doesn’t change.

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MCQs No. 58

In simple pendulum, suspension point shifts upward. Period:
a) Decreases
b) Increases
c) Unchanged
d) Goes to zero

Correct Answer: a. Decreases
Explanation: Effective length decreases.

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MCQs No.59

A simple pendulum is suspended from the roof of a lift. When the lift is moving downward with an acceleration “a” then the net acceleration is:
a. g
b. g – a
c. g + a
d) $\sqrt{g^2+a^2}$

The Correct Answer is option b. g – a 

Explanation:

In downward direction net acceleration decreases.

When the lift accelerates downward, the effective gravitational acceleration acting on the pendulum decreases.

$$a^{\mathrm{\prime }}=g-a$$

Since the time period of a simple pendulum depends on effective gravity, it becomes:

$$T=2\pi \sqrt{\frac{L}{g-a}}$$$$\boxed{{\displaystyle \text{Thus, }{a}^{\mathrm{\prime }}=g-a}}$$

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MCQs No.60
How much will be the length of a simple pendulum if its time period is one second?
a. 2.5 m
b. 0.25 m
c. 25 m
d. 0.025 m

The Correct Answer is option b. 0.25 m

Explanation:

The time period of a simple pendulum is given by

$$T = 2\pi \sqrt{\frac{l}{g}}$$

Given:
$T = 1 \,\text{s}$  and $g \approx 9.8 \,\text{m/s}^2$ 

$$1 = 2\pi \sqrt{\frac{l}{9.8}}$$ $$\sqrt{\frac{l}{9.8}} = \frac{1}{2\pi}$$ $$\frac{l}{9.8} = \frac{1}{4\pi^2}$$$$l = \frac{9.8}{4\pi^2} \approx 0.25 \,\text{m}$$

Hence, the correct answer is option b. 0.25 m.

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MCQs No.61

Damping is ____________________ related to amplitude of oscillations?
a. directly
b. inversely
c. linearly
d. none of these

The Correct Answer is option b. inversely 

Explanation:

Damping causes a continuous loss of energy in an oscillating system due to resistive forces like friction or air resistance. As damping increases, more energy is lost per oscillation, which results in a decrease in the amplitude of oscillations.

Thus, amplitude decreases when damping increases, meaning damping is inversely related to the amplitude of oscillations.

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MCQs No. 61

The maximum acceleration in SHM is:
a) ω²A
b) ωA
c) A/ω
d) A/ω²

Correct Answer: a. ω²A
Explanation: $a_{\max} = ω²A$.

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MCQs No. 62

Phase difference between displacement and acceleration is:
a) π
b) π/2
c) 0
d) 2π

Correct Answer: a. π
Explanation: Acceleration is opposite sign of x.

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MCQs No. 63

In SHM, total energy is proportional to:
a) Amplitude
b) Frequency
c) Amplitude²
d) Mass only

Correct Answer: c. Amplitude²
Explanation: E ∝ A².

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MCQs No. 64

A mass-spring system has period T. If k is increased four times, new T is:
a) T/2
b) T/4
c) 2T
d) Same

Correct Answer: a. T/2
Explanation: $T=2\pi\sqrt{m/k}$

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MCQs No.65

If the length of a simple pendulum is halved and its mass is doubled, then its time period will:

a. $\dfrac{T}{\sqrt{2}}$

b. $\sqrt{2}T$

c. Remains constant

d. none of these

Correct Answer is option a. $\dfrac{T}{\sqrt{2}}$

Explanation:
The time period of a simple pendulum is given by:

$$T = 2\pi \sqrt{\frac{\ell}{g}}$$

Time period is independent of mass.

$$\frac{T_2}{T_1} = \sqrt{\frac{\ell_2}{\ell_1}}$$ $$\ell_2 = \frac{\ell_1}{2} \Rightarrow T_2 = \sqrt{\frac{1}{2}}\,T_1 = \frac{T_1}{\sqrt{2}}$$

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MCQs No.66

The time period of a simple pendulum is 2 seconds. If the length of the pendulum is increased four times, the new time period will be:

a. 16 s
b. 12 s
c. 8 s
d. 4 s

Correct Answer is option d. 4 s

Explanation:

$$\frac{T_2}{T_1} = \sqrt{\frac{\ell_2}{\ell_1}} = \sqrt{4} \Rightarrow T_2 = 2T_1$$ $$T_2 = 2 \times 2 = 4 \text{ s}$$

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MCQs No.67

When the amplitude of a wave is doubled, its energy becomes:

a. Double
b. Four times
c. One half
d. None of these

Correct Answer is option b. Four times

Explanation:
Energy of a wave is proportional to the square of amplitude:

$$E \propto A^2$$

If amplitude is doubled:

$$E_2 = (2A)^2 = 4E_1$$

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MCQ No. 68

SHM is the projection of:
a) Circular motion
b) Rectilinear motion
c) Projectile motion
d) Random motion

Correct Answer: a. Circular motion

Explanation: 

Uniform circular → SHM projection.

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MCQs No. 69

If at x=A/2, KE =? of total energy.
a) 1/4 E
b) 1/2 E
c) 3/4 E
d) 1/3 E

Correct Answer: a. 1/4 E

Explanation: 

KE = E – PE, PE = 1/4E.

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MCQs No. 70

In SHM, x varies:
a) Linearly
b) Sinusoidally
c) Square
d) Randomly

Correct Answer: b. Sinusoidally

Explanation: 

By definition.

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MCQs No.71

For what displacement does the potential energy of a particle in SHM become one-fourth of its maximum value?

a. $x = x_0$

b. $x = \dfrac{x_0}{2}$

c. $x = \dfrac{x_0}{4}$

d. $x = \sqrt{x_0^2 - 2x}$

Correct Answer is option b. $x = \dfrac{x_0}{2}$

Explanation:
Potential energy in SHM:

$$\text{P.E} = \frac{1}{2}kx^2$$

Maximum (total) energy:

$$\text{T.E} = \frac{1}{2}k x_0^2$$

Given:

$$\text{P.E} = \frac{1}{4}\text{T.E}$$

$$\frac{1}{2}kx^2 = \frac{1}{4}\left(\frac{1}{2}k x_0^2\right) \Rightarrow x^2 = \frac{x_0^2}{4} \Rightarrow x = \frac{x_0}{2}$$

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MCQs No.72

If the potential energy of a vibrating particle is one-fourth of its total energy, the ratio of its displacement to amplitude is:

a. 1 : 1
b. 1 : 2
c. 1 : 4
d. 4 : 1

Correct Answer is option b. 1 : 2

Explanation:

$$\text{P.E} = \frac{1}{2}kx^2, \quad \text{T.E} = \frac{1}{2}k x_0^2$$

Given:

$$\text{P.E} = \frac{1}{4}\text{T.E} \Rightarrow x = \frac{x_0}{2}$$

Thus,

$$x : x_0 = 1 : 2$$

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MCQs No.73

A particle executing SHM has total energy E. If its displacement is reduced to half of the amplitude, its kinetic energy will be:

a. $\dfrac{E}{2}$
b. $\dfrac{E}{4}$
c. $\dfrac{3}{4}E$
d. $\sqrt{\dfrac{3}{4}E}$

Correct Answer: c. $\dfrac{3}{4}E$

Explanation:

Total energy of SHM:

$$E = \frac{1}{2}k x_0^2$$

Kinetic enerA particle executing SHM has total energy E. If its displacement is reduced to half of the amplitude, its kinetic energy will be:

a. $\dfrac{E}{2}$

b. $\dfrac{E}{4}$

c. $\dfrac{3}{4}E$

d. $\sqrt{\dfrac{3}{4}E}$

Correct Answer is option c. $\dfrac{3}{4}E$

Explanation:

Total energy of SHM:

$$E = \frac{1}{2}k x_0^2$$

Kinetic energy at displacement $x$:

$$\text{K.E} = \frac{1}{2}k(x_0^2 - x^2)$$

Given:

$$x = \frac{x_0}{2}$$$$\text{K.E} = \frac{1}{2}k\left(x_0^2 - \frac{x_0^2}{4}\right) = \frac{1}{2}k\left(\frac{3x_0^2}{4}\right)$$$$\text{K.E} = \frac{3}{4}\left(\frac{1}{2}k x_0^2\right) = \frac{3}{4}E$$

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MCQ No. 74

In a simple pendulum the motion is:
a) SHM for small angles
b) Not SHM
c) Projectile motion
d) Circular motion

Correct Answer: a. SHM for small angles

Explanation: 

Only small angles satisfy linear restoring force.

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MCQs No. 75

At extreme position in SHM, velocity is:
a) Maximum
b) Zero
c) Average
d) Half

Correct Answer: b. Zero

Explanation: 

Turning point.

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MCQ No. 76

The period of SHM is least dependent on:
a) Amplitude (small)
b) Mass
c) Gravity
d) Spring constant

Correct Answer is option a. Amplitude (small)

Explanation: 

Small amplitude doesn’t affect period.

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MCQs No.77

When the kinetic energy of a particle executing SHM is maximum, its:

a. Potential energy is zero
b. Acceleration is zero
c. Restoring force is zero
d. All of the above

Correct Answer is option d. All of the above

Explanation:
In SHM, kinetic energy is maximum at the mean position. At this position:

• Displacement is zero
• Potential energy is zero
• Acceleration is zero
• Restoring force is zero

Hence, all the given statements are correct.

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MCQs No.78

The oscillations that occur in the absence of an external force are called:

a. Free oscillations
b. Forced oscillations
c. Driven oscillations
d. Damped oscillations

Correct Answer is option a. Free oscillations

Explanation:
When a body oscillates without the action of any external periodic force, the oscillations are known as free oscillations.

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MCQs No.79

The uniform heating and cooking of food in a microwave oven is an example of:

a. Resonance
b. Specific heat
c. Damped oscillations
d. None of these

Correct Answer is option a. Resonance

Explanation:

Resonance is the phenomenon in which a system oscillates with large amplitude when the frequency of the applied force matches the natural frequency of the system. In a microwave oven, resonance helps in the efficient and uniform heating of food.

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MCQs No. 80

Energy in SHM changes between:
a) KE and PE
b) KE and temperature
c) PE and heat
d) KE and sound

Correct Answer: a. KE and PE

Explanation: 

Energy interchanges.

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MCQs No. 81

If mass doubles in mass-spring SHM, period:
a) Doubles
b) Quadruples
c) Increases by √2
d) Decreases

Correct Answer: c. Increases by √2

Explanation: 

$T\propto\sqrt{m}$

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MCQs No. 82

The mean position in SHM is also called:
a) Equilibrium
b) Extreme
c) Turning point
d) Zero velocity point

Correct Answer: a. Equilibrium
Explanation: By definition.

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MCQs No.83

In damped harmonic oscillation, which one decreases?
a. Amplitude of vibration
b. Energy of vibration
c. Both amplitude and energy
d. Neither amplitude nor energy

The Correct Answer is option c. Both amplitude and energy

Explanation:
Because energy losses in damped oscillation.

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MCQs No.84

Sharpness of resonance is: 
a. Directly proportional to damping force
b. Inversely proportional to damping force
c. Equal to square of damping force
d. Equal to square root of damping force.

The Correct Answer is option b. Inversely proportional to damping force

Explanation:

Sharpness of resonance depends on how small the damping force is in an oscillating system. When damping is small, the resonance curve is sharp; when damping increases, the resonance curve becomes broader.

Hence, sharpness of resonance is inversely proportional to the damping force.

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MCQs No.85

Electrical resonance is observed in:
a. Swing
b. Simple pendulum
c. Radio
d. None of these

The Correct Answer is option c. Radio 

Explanation:

Electrical resonance occurs in electrical circuits (such as LC or RLC circuits) when the frequency of the applied alternating current matches the natural frequency of the circuit. This principle is used in radio receivers to select and tune a desired station.

Hence, electrical resonance is observed in a radio.

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MCQs No. 86

If amplitude is reduced by half, total energy becomes:
a) Quarter
b) Half
c) Double
d) Same

Correct Answer: a. Quarter
Explanation: E ∝ A².

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MCQs No. 87

Oscillations with constant amplitude have:
a) No damping
b) Maximum damping
c) Zero frequency
d) Infinite energy

Correct Answer: a. No damping
Explanation: Damping reduces amplitude.

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MCQs No.88

MRI works on the principle of:

a. Beats
b. Interference
c. Resonance
d. Standing waves

Correct Answer: c. Resonance

Explanation:

MRI (Magnetic Resonance Imaging) works on the principle of magnetic resonance, in which nuclei absorb energy when the frequency of the applied electromagnetic radiation matches their natural frequency.

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MCQs No.89

The time periods of the same simple pendulum at Murree (M) and Karachi (K) are related as:

a. $T_K = T_M$
b. $T_K > T_M$
c. $T_K < T_M$
d. $2T_K = 3T_M$

Correct Answer: c. $T_K < T_M$

Explanation:
The time period of a simple pendulum is:

$$T \propto \frac{1}{\sqrt{g}}$$

At higher altitude (Murree), the value of $g$ is smaller, so the time period is greater.
At lower altitude (Karachi), $g$ is larger, so the time period is smaller.

Therefore:

$$T_K < T_M$$

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MCQs No.90

The even heating and cooking of food in a microwave oven is an example of:

a. Simple harmonic motion
b. Resonance
c. Adiabatic process
d. None of these

Correct Answer: b. Resonance

Explanation:
Microwaves produce resonance in water molecules, causing them to vibrate rapidly and generate heat, which results in uniform cooking of food.

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MCQs No. 91

Frequency of SHM increases if:
a) k increases
b) m increases
c) g decreases
d) A increases

Correct Answer: a. k increases
Explanation: ω²=k/m.

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MCQ No. 92

Maximum velocity in a pendulum occurs at:
a) Lowest point
b) Highest point
c) Any point
d) Zero

Correct Answer: a. Lowest point
Explanation: Potential least, kinetic max.

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MCQs No.93

The motion is repeated in equal interval of time is called ……….. motion .
a. Circular
b. Harmonic
c. periodic
d. None of these
 

The Correct Answer is option c. periodic 

Explanation:
The motion which repeats itself in equal interval of time is called periodic motion.

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MCQs No.94

To make the frequency double of a spring oscillation, we have to………
a. Double the mass
b. Half the mass
c. Quadrupole the mass
d. Reduce the mass to one fourth

The Correct Answer is option d. Reduce the mass to one fourth

Explanation:
The frequency of mass attached to the spring will be double when the mass is reduced one fourth. According to formula of frequency of mass attached to the spring is inversely proportional the square root of mass.

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MCQs No.95

In an isolated system the total energy of vibrating mass and spring is__________.
a. Constant
b. High
c. Low
d. Variable

The Correct Answer is option a. Constant

Explanation:
As in isolated system there is no external force acting on the system to change the energy of the system.

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MCQs No.96

If the period of oscillation of mass (M) suspended from a spring is 2 seconds, the period of mass 4M will be:
a. 4 seconds
b. 3 seconds
c. 2 seconds
d. 1 second

The Correct Answer is option a. 4 seconds 

Explanation:

$T\propto\sqrt{m}$

As the time period of mass attached to the spring is proportional to the square root of mass and by increasing the mass 4 times the time period increases by 2 times.

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MCQs No.97

The time period of simple pendulum dese not depends upon __________.
a. gravity
b. length
c. mass
d. frequency

The Correct Answer is option c. mass

Explanation:
The time period of simple pendulum is independent of mass of the bob.

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MCQs No.98

A Spring of Spring Constant k is cut in to three equal parts, the spring constant of each part will be _________________.
a. 3k
b. 9k
c. k
d. k/3

The Correct Answer is option c. k

Explanation:
The value of K is inversely proportional to length.

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MCQs No.99

Mathematically simple harmonic motion can be expressed as __________.
a. a α x
b. a α x²
c. a α -x²
d. a α -x

The Correct Answer is option d. a α -x 

Explanation:
By definition of SHM.

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MCQs No.100

If frequency of a simple pendulum is 2Hz then time period is:
1 sec
2 sec
0.5 sec
1.5sec

The Correct Answer is option c. 0.5 sec 

Explanation:
As f= 1/T =1/2 = 0.5 Hz.

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FBISE past model and exam‑style MCQson oscillations/SHM

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MCQ No. 1

In a simple pendulum experiment, the length of the pendulum is 50.0 cm. If the acceleration due to gravity is 9.8 m/s², what is the frequency of the pendulum’s oscillation?

a) 0.22 Hz
b) 0.50 Hz
c) 0.25 Hz
d) 0.10 Hz

Correct Answer: a) 0.22 Hz
Explanation:

$$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.50}{9.8}} \approx 1.43\,\text{s}$$$$f = \frac{1}{T} \approx \frac{1}{1.43} \approx 0.70\,\text{Hz}\approx 0.22\,\text{Hz}$$

(Frequency of pendulum is inverse of time period)

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MCQ No. 2

In a simple pendulum experiment, the percentage error in measuring the length L is 0.1 % and in measuring time T for oscillations is 2 %. What is the approximate percentage error in g calculated from the data?

a) 4.1 %
b) 3.1 %
c) 5 %
d) 2.1 %

Correct Answer: a) 4.1 %
Explanation:
Percent error in g from pendulum formula involves doubling the percent error in T and adding percent error in L.

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MCQ No. 3

The time for 10 oscillations of a simple pendulum is measured as 20.0 s, and the length of the pendulum is 1.00 m. What is the experimental value of g?

a) 9.8 m/s²
b) 9.6 m/s²
c) 9.5 m/s²
d) 10.0 m/s²

Correct Answer: a) 9.8 m/s²
Explanation:
Time period per oscillation = 20 s ÷ 10 = 2.0 s, so

$$T=2\pi\sqrt{\frac{L}{g}} ⇒ g=\frac{4\pi^2L}{T^2}= \frac{4\pi^2(1)}{4}≈9.8$$

(Derived from simple pendulum formula)

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MCQ No. 4

A mass‑spring system oscillates with angular frequency 5 rad/s and amplitude 0.10 m. What is the total energy of the system (in J) if the mass is 0.50 kg?

a) 0.125 J
b) 0.312 J
c) 0.250 J
d) 0.500 J

Correct Answer: c) 0.250 J
Explanation:

$$E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}(0.50)(5^2)(0.10^2)=0.25\,\text{J}$$

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MCQ No. 5

The displacement of a SHM particle is given by

$$x = 0.15\cos(10t)$$

What is the maximum acceleration?

a) 15 m/s²
b) 22.5 m/s²
c) 10 m/s²
d) 7.5 m/s²

Correct Answer: b) 22.5 m/s²
Explanation:

$$a_{\max} = \omega^2A = (10^2)(0.15)=100×0.15=15 m/s^2$$

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MCQ No. 6

The period of a simple pendulum is 2 s. What is its frequency?

a) 2 Hz
b) 1 Hz
c) 0.5 Hz
d) 4 Hz

The Correct Answer is option c. 0.5 Hz

Explanation:

$$f = \frac{1}{T} = \frac{1}{2} = 0.5\,\text{Hz}$$

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MCQ No. 7

A particle performing SHM has total energy E. What is its kinetic energy at displacement $x = \frac{A}{2}$ ?

a) $\frac{E}{4}$

b) $\frac{E}{2}$

c) $\frac{3E}{4}$

d) $E$

The Correct Answer is option a. $\frac{E}{4}$

Explanation:

$$K.E = E - P.E$$

$$P.E = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{4}E$$

$$K.E = E - \frac{3}{4}E = \frac{1}{4}E$$

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