Unit-7 Oscillation – Complete Exercise Solutions | Class 11 Physics | FBISE

Unit-7 Oscillation – Complete Exercise Solutions | Class 11 Physics | KPK Text Board, Peshawar | FBISE

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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This post provides a complete and fully solved exercise of Unit-7: Oscillation for Class 11 Physics, strictly according to the KPK Textbook Board / FBISE syllabus. It includes all MCQs, short-answer questions, comprehensive (long) questions, and numerical problems, solved step-by-step using standard board-style notation, formulas, and clear explanations.

Whether you are preparing for annual board examinations, chapter tests, or entry tests, this guide is designed to be concept-clear, exam-oriented, and high-scoring, helping you master the chapter with confidence.

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1. Choose the best possible answer of the following MCQs.

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MCQs No. 1

1. Tuning of a radio set is an example of:

a. Mechanical resonance
b. Musical resonance
c. Electrical resonance
d. Free vibrations

Correct Answer: c. Electrical resonance
Explanation: A radio tuner selects a particular frequency using electrical resonance in LC circuits.

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2. The heating and cooking of food evenly by a microwave oven is an example of:

a. SHM
b. Resonance
c. Damped oscillation
d. Free oscillation

Correct Answer: b. Resonance

Explanation: Microwaves cause resonance in water molecules, producing uniform heating.

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3. The time period of the same pendulum at Karachi and Murree 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: b. $T_K>T_M$ 

Explanation: Gravitational acceleration (g) decreases at higher altitudes; Murree is higher than Karachi → $g_{M} < g_{K}$, so $T_M > T_K$.

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4. In an isolated system, the total energy of a vibrating mass and spring is:

a. Variable
b. Low
c. High
d. Constant

Correct Answer: d. Constant

Explanation: In ideal SHM, mechanical energy (KE + PE) remains constant in the absence of damping.

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5. While deriving the equation of time period for a simple pendulum, which quantity should be kept small?

a. Length of the pendulum
b. Amplitude
c. Mass of the pendulum
d. Gravitational acceleration g

Correct Answer: b. Amplitude
Explanation: Small angular displacement (amplitude) ensures sinθ ≈ θ, which is required for SHM derivation.

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6. If the period of oscillation of mass $M$ suspended from a spring is 2 s, then the period of mass $4M$ will be:

a. 1 s
b. 2 s
c. 3 s
d. 4 s

Correct Answer: d. 4 s
Explanation: $T \propto \sqrt{m} \Rightarrow T' = \sqrt{4} T = 2 × 2 = 4$s.

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7. The time period of a simple pendulum is 2 s. If its length is increased by 4 times, then its period becomes:

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

Correct Answer: d. 4 s
Explanation: $T \propto \sqrt{L} \Rightarrow T' = \sqrt{4} × 2 = 2 × 2 = 4$s.

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8. To make the frequency of a spring oscillation double, we have to:

a. Reduce the mass to one fourth
b. Quadruple the mass
c. Double the mass
d. Halve the mass

Correct Answer: a. Reduce the mass to one fourth
Explanation: $f \propto 1/\sqrt{m}$

To double f: $m\to m\mathrm{/}\mathrm{4 }$

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9. The restoring force of SHM is maximum when the particle is at:

a. Maximum displacement
b. Halfway between mean and extreme
c. Crossing the mean position
d. At rest

Correct Answer: a. Maximum displacement
Explanation: $F = -kx$. Maximum displacement → maximum x → maximum restoring force.

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10. Two springs of spring constants $k_1$ and $k_2$are joined in series. The effective spring constant is:

a. $(k_1 + k_2)/2$ 

b. $k_1 + k_2$

c. $\dfrac{k_1 k_2}{k_1 + k_2}$

d. $\sqrt{k_1 k_2}$

Correct Answer: c. ${\displaystyle \frac{k_1{k}_2}{k_1+k_2}}$
Explanation: For springs in series: $\dfrac{1}{k_\text{eff}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \Rightarrow k_\text{eff} = \dfrac{k_1 k_2}{k_1 + k_2}$.

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If you want, I can polish the n

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CONCEPTUAL QUESTIONS

Give short response to the following questions

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1. Give two applications in which resonance plays an important role.

Answer:

• Tuning a radio to select a particular frequency.

• Microwave ovens for heating food.
  (Other examples: musical instruments, bridges responding to wind/traffic.)

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2. What happens to the time period of a simple pendulum if its length is doubled?

Answer:
Time period increases by factor $\sqrt{2}$.

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

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3. What will be the frequency of a simple pendulum if its length is 1 m?

Answer:

$$T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.01 \, \text{s} f = 1/T \approx 0.5 \, \text{Hz}$$

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4. Give one practical example each of free and forced oscillation.

Answer:

• Free oscillation: A simple pendulum swinging after being released.
• Forced oscillation: Child on a swing pushed periodically by a parent.

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5. How can you compare the masses of two bodies by observing their frequencies of oscillation when supported by a spring?

Answer:

$$f\propto \frac{1}{\sqrt{m}}\text{<span class="mord text"><br></span><span class="mord text"><span class="mord"><br></span></span><span class="mord text"><span class="mord">Lower frequency → larger mass, </span></span><span class="mord text"><span class="mord">higher frequency → smaller mass.</span></span>}$$

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6. A wire hangs from the top of a dark high tower, so that the top of the tower is not visible. How would you determine the height of that tower?

Answer:

Cause the wire to oscillate as a simple pendulum.

Measure the time period $T$.

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

to calculate the length L = height of the tower.

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7. Why in SHM the acceleration is zero when the velocity is greatest?

Answer:

Acceleration $a=-{\omega }^{2}x$  depends on displacement.

Maximum velocity occurs at mean position $x=0$  → $a=0$.

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8. What is the total distance covered by a simple harmonic oscillator in a time equal to its period if the amplitude is $A$?

Answer:

One complete oscillation: moves from $+A\to -A\to +A$ 

Total distance $=4A$.

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9. What happens to the frequency of a simple pendulum as its oscillations die down from large amplitude to small?

Answer:

• For small amplitudes, SHM approximation applies: frequency remains practically constant.
• Slightly larger amplitudes → frequency decreases a little, but effect is negligible.

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10. A singer, holding a note of right frequency, can shatter a glass. Explain.

Answer:

• If the singer’s note matches the natural frequency of the glass → resonance occurs.
• Amplitude of glass vibrations increases → glass may break.

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COMPREHENSIVE QUESTIONS

Give extended response to the following question

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1. Show that motion of a mass attached with a spring executes SHM

Answer/Solution:

Consider a mass $m$ attached to a spring of spring constant $k$.

If the mass is displaced by $x$ from equilibrium, the restoring force is:

$$F = -kx$$

By Newton’s 2nd law:

$$F = ma \Rightarrow m \frac{d^2x}{dt^2} = -kx \Rightarrow \frac{d^2x}{dt^2} + \frac{k}{m}x = 0$$

This is the defining equation of SHM:

$$\frac{d^2x}{dt^2} + \omega^2 x = 0 \quad \text{where} \quad \omega = \sqrt{\frac{k}{m}}$$

Hence, the mass executes simple harmonic motion with angular frequency

$\omega =\sqrt{k\mathrm{/}m}$

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2. Prove that the projection of a body moving in a circle describes SHM

Solution:

Consider a particle moving in a circle of radius $R$ with angular velocity $\omega$.

Let $x$ be the projection on the horizontal diameter.

$$x = R \cos \theta = R \cos(\omega t)$$

Differentiating twice w.r.t time:

$$\frac{d^2x}{dt^2} = -\omega^2 R \cos(\omega t) = -\omega^2 x$$

 This is the SHM equation: $a=-{\omega }^{2}x$ 

Therefore, the projection of circular motion is simple harmonic motion.

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3. Show that energy is conserved in case of SHM

Solution:

For mass $m$  on a spring: displacement $x=A\cos (\omega t)$ 

Velocity: $v=\frac{dx}{dt}=-\omega A\sin (\omega t)$ 

Kinetic energy:

$$KE = \frac{1}{2} m v^2 = \frac{1}{2} m (\omega^2 A^2 \sin^2 \omega t)$$

Potential energy in spring:

$$PE = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \cos^2 \omega t$$

Total energy:

$$E = KE + PE = \frac{1}{2} k A^2 (\sin^2 \omega t + \cos^2 \omega t) = \frac{1}{2} k A^2$$

Since $E$ is constant, total energy is conserved in SHM.

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4. Differentiate free and forced oscillations

Answer:

Free oscillations

Free oscillation occurs when a system oscillates naturally due to an initial displacement, without continuous external influence. Its frequency is equal to the natural frequency of the system, and the amplitude remains constant if there is no damping. An example is a simple pendulum released from rest. 

Forced oscillations

In contrast, forced oscillation occurs when an external periodic force continuously drives the system. The frequency of oscillation in this case equals the frequency of the driving force, and the amplitude depends on the driving frequency. A practical example is a child on a swing being pushed periodically.

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5. What is resonance? Give three applications in daily life

Answer:

Resonance: When the frequency of an external periodic force matches the natural frequency of a system, the amplitude becomes maximum.

Applications:

1. Tuning a radio to a desired station.

2. Microwave oven heating food evenly.

3. Musical instruments producing loud sound at natural frequencies.

(Other examples: bridges collapsing due to wind, resonance in strings, and glass shattering by voice.)

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6. Derive equations for kinetic and potential energy of a body of mass m executing SHM

Solution:

Displacement in SHM: $x=A\cos (\omega t)$

Velocity: $v=\frac{dx}{dt}=-\omega A\sin (\omega t)$

Kinetic energy:

$$KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$$

Potential energy:

$$PE = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \cos^2(\omega t) = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t)$$

Total energy:

$$E = KE + PE = \frac{1}{2} m \omega^2 A^2 = \text{constant}$$

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7. Explain what is meant by damped oscillations

Answer:

Damped oscillation occurs when resistive forces (like friction or air resistance) act on an oscillating system.

These forces reduce amplitude gradually over time.

Types of damping:

1. Light damping: oscillations continue, slowly decreasing amplitude.

2. Critical damping: system returns to equilibrium without oscillating.

3. Heavy damping: system returns slowly without completing full oscillation.

Example: Car shock absorbers reduce oscillations for smooth ride.

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4. Numerical Problems:

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Numerical No. 1

Question:
A force of 0.4 N is required to displace a body attached to a spring through 0.1 m from its mean position. Calculate the spring constant of spring.

Given Data:

$$F = 0.4 \, \text{N}, \quad x = 0.1 \, \text{m}$$

To Find:

$$k \, (\text{spring constant})$$

Solution:

$$F = kx \Rightarrow k = \frac{F}{x} = \frac{0.4}{0.1} = 4 \, \text{N/m}$$

Answer:

$$k = 4 \, \text{N/m}$$

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Numerical No. 2

Question:
A pendulum clock keeps perfect time at $g = 9.8 \, \text{m/s²}$. When moved to a higher altitude, it loses 80.0 s per day. Find the value of $g$ at the new location.

Given Data:

$$g_0 = 9.8 \, \text{m/s²}, \quad \Delta T = 80 \, \text{s/day}, \quad T_\text{day} = 86400 \, \text{s}$$

To Find:

$$g_\text{new}$$

Solution:

$$\frac{\Delta T}{T} = \frac{80}{86400} = 0.0009259$$$$T' = T \sqrt{\frac{g_0}{g}} \Rightarrow \frac{T' - T}{T} = \sqrt{\frac{g_0}{g}} - 1$$
$$0.0009259 = \sqrt{\frac{9.8}{g}} - 1 \Rightarrow \sqrt{\frac{9.8}{g}} = 1.0009259$$
$$\frac{9.8}{g} = 1.001852 \Rightarrow g \approx 9.782 \, \text{m/s²}$$

Answer:

$$g \approx 9.782 \, \text{m/s²}$$

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Numerical No. 3

Question:
Calculate the length of a second pendulum having time period 2 s at a place where $g = 9.8 \, \text{m/s²}$.

Given Data:

$$T = 2 \, \text{s}, \quad g = 9.8 \, \text{m/s²}$$

To Find:

$$L \, (\text{length of pendulum})$$

Solution:

$$T = 2 \pi \sqrt{\frac{L}{g}} \Rightarrow L = \frac{T^2 g}{4 \pi^2} = \frac{4 × 9.8}{39.48} \approx 0.994 \, \text{m}$$

Answer:

$$L \approx 0.994 \, \text{m}$$

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Numerical No. 4

Question:
A body of mass $m$ suspended from a spring with force constant $k$  vibrates with frequency $f_1$ . When the spring is cut into half and the same mass is suspended from one half, the frequency becomes $f_2$. Find $f_2/f_1$⁻¹

Given Data:

$$\text{Original frequency } f_1, \quad \text{Spring halved → } k \to 2k$$

To Find:

$$f_2/f_1$$

Solution:

$$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}} \Rightarrow f_2 = \frac{1}{2\pi} \sqrt{\frac{2k}{m}} = \sqrt{2} f_1$$

Answer:

$$f_2/f_1 = \sqrt{2}$$

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Numerical No. 5

Question:
A mass at the end of a spring describes SHM with $T=0.40\text{s}$. Find acceleration when displacement is 0.04 m.

Given Data:

$$T=0.40\text{s},$$

$$x=0.04\text{m}$$

To Find:

$$a(\text{acceleration})$$

Solution:

$$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.40} = 15.708 \, \text{rad/s}$$$$a = - \omega^2 x = - (15.708)^2 × 0.04 \approx -9.86 \, \text{m/s²}$$

Answer:

$$a \approx -9.86 \, \text{m/s²}$$

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Numerical No. 6

Question:
A block weighing 4.0 kg extends a spring by 0.16 m from its unstretched position. The block is removed and a 0.50 kg body is hung from the same spring. Find the period of vibration.

Given Data:

$$m_1 = 4.0 \, \text{kg}, \quad x_1 = 0.16 \, \text{m}, \quad m_2 = 0.50 \, \text{kg}$$

To Find:

$$T \, (\text{period for } m_2)$$

Solution:

$$k = \frac{m_1 g}{x_1} = \frac{4 × 9.8}{0.16} = 245 \, \text{N/m}$$$$T = 2 \pi \sqrt{\frac{m_2}{k}} = 2 \pi \sqrt{\frac{0.5}{245}} = 0.284 \, \text{s}$$

Answer:

$$T \approx 0.284 \, \text{s}$$

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Numerical No. 7

Question:
What should be the length of a simple pendulum whose time period is 1 s? Also find its frequency.

Given Data:

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

$$g=9.8\text{m/s²}$$

To Find:

$$L(\text{length})\; =\; ?$$

$$f(\text{frequency})\; =\; ?$$

Solution:

$$L = \frac{T^2 g}{4 \pi^2} = \frac{1 × 9.8}{39.48} \approx 0.248 \, \text{m}$$$$f = 1/T = 1 \, \text{Hz}$$

Answer:

$$L \approx 0.248 \, \text{m}, \quad f = 1 \, \text{Hz}$$cps = cycle per second

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Numerical No. 8

Question:
A spring, $k = 80 \, \text{N/m}$, supports a mass of 1 kg at rest. Find the distance by which the mass must be pulled down so that it passes the mean position with $v=1\text{m/s}$.

Given Data:

$$k=80\text{N/m},$$

$$m=1\text{kg},$$

$$v=1\text{m/s}$$

To Find:

$$x(\text{displacement})\; =\; ?$$

Solution:

$$E = \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \Rightarrow x = \sqrt{\frac{m v^2}{k}} = \sqrt{\frac{1 × 1^2}{80}} \approx 0.112 \, \text{m}$$

Answer:

$$x \approx 0.112 \, \text{m}$$

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Numerical No. 9

Question:
An 800 g body vibrates SHM with amplitude 0.30 m. The restoring force is 60 N at maximum displacement. Find: (i) $T$, (ii) $a$ at $x = 0.12$ m, (iii)  $v$, (iv) $KE$, (v) $PE$

Given Data:

$$m = 0.8 \, \text{kg}, \quad A = 0.3 \, \text{m}, \quad F_\text{max} = 60 \, \text{N}, \quad x = 0.12 \, \text{m}$$

To Find:

(i) $T=\; ?$ 

(ii) a at x =0.12 m

(iii) v=? 

(iv) KE =? 

(v) PE =?

Solution:
Spring constant: $k=F_{\text{max}}\mathrm{/}A=60\mathrm{/}0.3=200\text{N/m}$ 

(i) Time period: $T=2\pi \sqrt{m\mathrm{/}k}=2\pi \sqrt{0.8\mathrm{/}200}\approx 0.397\text{s}$

(ii) Acceleration: $a=kx\mathrm{/}m=200\times 0.12\mathrm{/}0.8=30\text{m/s²}$

(iii) Angular frequency: $\omega =\sqrt{k\mathrm{/}m}=\sqrt{200\mathrm{/}0.8}=15.81\text{rad/s}$

$$v = \omega \sqrt{A^2 - x^2} = 15.81 \sqrt{0.09 - 0.0144} = 4.35 \, \text{m/s}$$
(iv) Kinetic energy: $KE=0.5\times m\times v^2=0.5\times 0.8\times {4.35}^2\approx 7.57\text{J}$

(v) Potential energy: $PE=0.5\times k\times x^2=0.5\times 200\times {0.12}^2=1.44\text{J}$

Answer:

$$T \approx 0.397 \, \text{s}, \quad a = 30 \, \text{m/s²}, \quad v = 4.35 \, \text{m/s}, \quad KE = 7.57 \, \text{J}, \quad PE = 1.44 \, \text{J}$$

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Numerical No. 10

Question:
Find amplitude, frequency, time period, and displacement at $t=2.0\text{s}$ if $x = 0.25 \cos(\pi/8 t)$.

Given Data:

$$x = 0.25 \cos(\pi/8 t)$$

To Find:

$$A, f, T, x(t=2)$$

Solution:

Amplitude: $A=0.25\text{m}$

Angular frequency: $\omega =\pi \mathrm{/}8\text{rad/s}$ 

Frequency: $f=\omega \mathrm{/}2\pi =(\pi \mathrm{/}8)\mathrm{/}(2\pi )=1\mathrm{/}16=0.0625\text{Hz}$

Time period: $T=1\mathrm{/}f=16\text{s }$

Displacement at $t=2\text{s}$ :

$$x = 0.25 \cos(\pi/8 × 2) = 0.25 \cos(\pi/4) = 0.25 × 0.707 \approx 0.177 \, \text{m}$$

Answer:

$$A = 0.25 \, \text{m}, \quad f = 0.0625 \, \text{Hz}, \quad T = 16 \, \text{s}, \quad x(2s) \approx 0.177 \, \text{m}$$

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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