Oscillatory Motion – 100 Advanced & Numerical MCQs with Solutions | ECAT, MDCAT & Engineering Entry Tests

Oscillatory Motion – 100 Advanced & Numerical MCQs with Solutions | ECAT, MDCAT & Engineering Entry Tests

100 Advanced & Numerical MCQs (Level -2) on Oscillatory Motion (Oscillations), Physics (Unit-Wise MCQs Practice):


Whether you are preparing for Board Exams, school tests, college assessments, university exams, ECAT, MDCAT, NTS, PIEAS, GIKI, UET, FAST, NUST, or other competitive examinations, this comprehensive Oscillatory Motion (Oscillations) MCQs Collection is designed to strengthen your conceptual understanding, analytical thinking, and numerical problem-solving skills.

This chapter covers all the essential topics, including oscillatory motion, free oscillations, forced oscillations, damped oscillations, Simple Harmonic Motion (SHM), conditions for SHM, projection of uniform circular motion, displacement, amplitude, time period, frequency, angular frequency, phase, phase difference, equation of SHM (a = –ω²x), restoring force, spring-mass system, energy changes during SHM, kinetic and potential energy, conservation of mechanical energy, simple pendulum, time period of a pendulum, resonance, natural frequency, critical damping, damping factor, car suspension system, practical applications of damping, factors affecting the frequency of oscillations, graphs of SHM, and real-life applications of oscillatory motion.

The MCQs are organized into four progressive sections:

100 Basic MCQs Level – 1 (1–100)

Build a strong foundation with essential concepts, definitions, characteristics, equations, graphs, laws, formulas, and fundamental principles of oscillatory motion and Simple Harmonic Motion.

100 Advanced & Numerical MCQs (101–200)

Strengthen your calculation, analytical thinking, and application-based problem-solving skills through challenging numerical and conceptual questions on spring-mass systems, simple pendulums, angular frequency, resonance, damping, velocity, acceleration, energy, and oscillatory motion.

50 Higher-Order Thinking Skills (HOTS) MCQs (201–250)

Challenge yourself with conceptual reasoning, graph interpretation, real-life applications, assertion-reason questions, experimental situations, engineering applications, resonance phenomena, damping analysis, and competitive examination-level problems.

50 Challenging MCQs Quiz with Answers

A carefully selected collection of the 50 most important conceptual, numerical, and HOTS questions designed for quick revision, self-assessment, concept reinforcement, and complete exam preparation.

Every MCQ Includes:

✔ Four carefully designed answer options

✔ Instant correct answer

✔ Detailed concept-based explanation

✔ Exam-focused learning approach

This complete MCQ collection is ideal for Board Exams, FBISE, ECAT, MDCAT, NTS, PIEAS, GIKI, UET, FAST, NUST, and other competitive entrance examinations. By practicing these 250 carefully selected MCQs along with the Top 50 Challenging MCQs Quiz, you will develop a thorough understanding of Oscillatory Motion and Simple Harmonic Motion, strengthen your numerical problem-solving abilities, improve conceptual clarity and logical reasoning, and gain the confidence needed to excel in objective-type Physics examinations and competitive entrance tests.

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    • Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)

    MCQ No. 101

    A particle performs SHM with a time period of 4 s. Its frequency is:

    a) 0.20 Hz

    b) 0.25 Hz

    c) 2 Hz

    d) 4 Hz

    Correct Answer: b) 0.25 Hz

    Explanation: Frequency is the reciprocal of the time period.

    f=1T=14=0.25 Hzf=\frac{1}{T}=\frac{1}{4}=0.25\text{ Hz}

    MCQ No. 102

    The frequency of an oscillator is 10 Hz. Its time period is:

    a) 0.01 s

    b) 0.1 s

    c) 10 s

    d) 20 s

    Correct Answer: b) 0.1 s

    Explanation:

    T=1f=110=0.1 sT=\frac{1}{f}=\frac{1}{10}=0.1\text{ s}

    MCQ No. 103

    The angular frequency of an oscillator having frequency 5 Hz is:

    a) 5π rad s⁻¹

    b) 10π rad s⁻¹

    c) 20π rad s⁻¹

    d) 25π rad s⁻¹

    Correct Answer: b) 10π rad s⁻¹

    Explanation:

    ω=2πf=2π(5)=10π rad s−1\omega=2\pi f=2\pi(5)=10\pi\text{ rad s}^{-1}

    MCQ No. 104

    An oscillator has an angular frequency of 8π rad s⁻¹. Its frequency is:

    a) 2 Hz

    b) 4 Hz

    c) 8 Hz

    d) 16 Hz

    Correct Answer: b) 4 Hz

    Explanation:

    f=ω2π=8π2π=4 Hzf=\frac{\omega}{2\pi}=\frac{8\pi}{2\pi}=4\text{ Hz}

    MCQ No. 105

    The time period of an oscillator having angular frequency 20π rad s⁻¹ is:

    a) 0.05 s

    b) 0.10 s

    c) 0.20 s

    d) 0.50 s

    Correct Answer: b) 0.10 s

    Explanation:

    T=2πω=2π20π=0.10 sT=\frac{2\pi}{\omega} =\frac{2\pi}{20\pi} =0.10\text{ s}

    MCQ No. 106

    The amplitude of a particle executing SHM is 8 cm. The maximum displacement from the mean position is:

    a) 4 cm

    b) 8 cm

    c) 16 cm

    d) 32 cm

    Correct Answer: b) 8 cm

    Explanation: The amplitude is the maximum displacement from the equilibrium position.

    MCQ No. 107

    A particle completes 300 oscillations in 2 minutes. Its frequency is:

    a) 1.5 Hz

    b) 2 Hz

    c) 2.5 Hz

    d) 3 Hz

    Correct Answer: c) 2.5 Hz

    Explanation:

    Time = 120 s

    f=300120=2.5 Hzf=\frac{300}{120}=2.5\text{ Hz}

    MCQ No. 108

    A particle has a frequency of 50 Hz. The number of oscillations completed in 20 s is:

    a) 250

    b) 500

    c) 1000

    d) 2000

    Correct Answer: c) 1000

    Explanation:

    N=ft=50×20=1000N=ft=50\times20=1000

    MCQ No. 109

    If the frequency of an oscillator doubles, its time period becomes:

    a) Double

    b) Four times

    c) Half

    d) Unchanged

    Correct Answer: c) Half

    Explanation: Frequency and time period are inversely proportional.

    MCQ No. 110

    The angular frequency of a particle is 100π rad s⁻¹. The frequency is:

    a) 25 Hz

    b) 50 Hz

    c) 75 Hz

    d) 100 Hz

    Correct Answer: b) 50 Hz

    Explanation:

    f=100π2π=50 Hzf=\frac{100\pi}{2\pi}=50\text{ Hz}

    MCQ No. 111

    A spring has a spring constant of 200 N m⁻¹. If it is stretched by 0.05 m, the restoring force is:

    a) 5 N

    b) 10 N

    c) 15 N

    d) 20 N

    Correct Answer: b) 10 N

    Explanation:

    F=kx=200(0.05)=10 NF=kx=200(0.05)=10\text{ N}

    MCQ No. 112

    A spring is stretched by 4 cm by a force of 8 N. The spring constant is:

    a) 100 N m⁻¹

    b) 150 N m⁻¹

    c) 200 N m⁻¹

    d) 250 N m⁻¹

    Correct Answer: c) 200 N m⁻¹

    Explanation:

    k=Fx=80.04=200 N m−1k=\frac{F}{x} =\frac{8}{0.04} =200\text{ N m}^{-1}

    MCQ No. 113

    The spring constant of a spring is 500 N m⁻¹. A force of 25 N produces an extension of:

    a) 2 cm

    b) 5 cm

    c) 8 cm

    d) 10 cm

    Correct Answer: b) 5 cm

    Explanation:

    x=Fk=25500=0.05 m=5 cmx=\frac{F}{k} =\frac{25}{500} =0.05\text{ m}=5\text{ cm}

    MCQ No. 114

    The time period of a spring-mass system is 2 s. Its frequency is:

    a) 0.25 Hz

    b) 0.5 Hz

    c) 1 Hz

    d) 2 Hz

    Correct Answer: b) 0.5 Hz

    Explanation:

    f=12=0.5 Hzf=\frac{1}{2}=0.5\text{ Hz}

    MCQ No. 115

    The time period of a spring-mass system is 0.5 s. Its angular frequency is:

    a) 2π rad s⁻¹

    b) 4π rad s⁻¹

    c) 8π rad s⁻¹

    d) 16π rad s⁻¹

    Correct Answer: b) 4π rad s⁻¹

    Explanation:

    ω=2πT=2π0.5=4π\omega=\frac{2\pi}{T} =\frac{2\pi}{0.5} =4\pi

    MCQ No. 116

    A particle has an amplitude of 10 cm. Its displacement can never be:

    a) 2 cm

    b) 8 cm

    c) 10 cm

    d) 15 cm

    Correct Answer: d) 15 cm

    Explanation: Displacement cannot exceed the amplitude.

    MCQ No. 117

    A pendulum completes one oscillation in 2 s. The number of oscillations in 1 minute is:

    a) 20

    b) 25

    c) 30

    d) 60

    Correct Answer: c) 30

    Explanation:

    N=602=30N=\frac{60}{2}=30

    MCQ No. 118

    A particle performs 600 oscillations in 5 minutes. Its frequency is:

    a) 1 Hz

    b) 2 Hz

    c) 3 Hz

    d) 4 Hz

    Correct Answer: b) 2 Hz

    Explanation:

    f=600300=2 Hzf=\frac{600}{300}=2\text{ Hz}

    MCQ No. 119

    If the frequency changes from 20 Hz to 40 Hz, the angular frequency becomes:

    a) Half

    b) Double

    c) Four times

    d) Unchanged

    Correct Answer: b) Double

    Explanation: Since

    ω=2πf\omega=2\pi f

    angular frequency is directly proportional to frequency.

    MCQ No. 120

    A spring constant increases from 100 N m⁻¹ to 400 N m⁻¹. The spring becomes:

    a) Softer

    b) Four times stiffer

    c) Half as stiff

    d) Unchanged

    Correct Answer: b) Four times stiffer

    Explanation: A larger spring constant indicates a stiffer spring.

    MCQ No. 121

    A particle has a frequency of 25 Hz. Its time period is:

    a) 0.02 s

    b) 0.04 s

    c) 0.25 s

    d) 0.40 s

    Correct Answer: b) 0.04 s

    Explanation:

    T=125=0.04 sT=\frac{1}{25}=0.04\text{ s}

    MCQ No. 122

    The angular frequency corresponding to 25 Hz is:

    a) 25π rad s⁻¹

    b) 50π rad s⁻¹

    c) 75π rad s⁻¹

    d) 100π rad s⁻¹

    Correct Answer: b) 50π rad s⁻¹

    Explanation:

    ω=2π(25)=50π\omega=2\pi(25)=50\pi

    MCQ No. 123

    A spring extends 0.10 m under a force of 40 N. Its spring constant is:

    a) 100 N m⁻¹

    b) 200 N m⁻¹

    c) 300 N m⁻¹

    d) 400 N m⁻¹

    Correct Answer: d) 400 N m⁻¹

    Explanation:

    k=400.10=400 N m−1k=\frac{40}{0.10}=400\text{ N m}^{-1}

    MCQ No. 124

    The amplitude of a particle is 6 cm. The distance between its two extreme positions is:

    a) 6 cm

    b) 8 cm

    c) 10 cm

    d) 12 cm

    Correct Answer: d) 12 cm

    Explanation: The distance between the two extreme positions is twice the amplitude.

    2A=2(6)=12 cm2A=2(6)=12\text{ cm}

    MCQ No. 125

    An oscillator completes 180 oscillations in 90 seconds. Its frequency is:

    a) 1 Hz

    b) 2 Hz

    c) 3 Hz

    d) 4 Hz

    Correct Answer: b) 2 Hz

    Explanation:

    f=18090=2 Hzf=\frac{180}{90}=2\text{ Hz}

    MCQ No. 126

    A body of mass 2 kg is attached to a spring of force constant 200 N m⁻¹. The time period of oscillation is:

    a) 0.20 s

    b) 0.40 s

    c) 0.63 s

    d) 1.26 s

    Correct Answer: c) 0.63 s

    Explanation:

    T=2πmk=2π2200=2π0.01=0.2π≈0.63 sT=2\pi\sqrt{\frac{m}{k}} =2\pi\sqrt{\frac{2}{200}} =2\pi\sqrt{0.01} =0.2\pi\approx0.63\text{ s}

    MCQ No. 127

    A spring has a force constant of 100 N m⁻¹. A 1 kg mass is attached to it. The angular frequency is:

    a) 5 rad s⁻¹

    b) 10 rad s⁻¹

    c) 15 rad s⁻¹

    d) 20 rad s⁻¹

    Correct Answer: b) 10 rad s⁻¹

    Explanation:

    ω=km=1001=10 rad s−1\omega=\sqrt{\frac{k}{m}} =\sqrt{\frac{100}{1}} =10\text{ rad s}^{-1}

    MCQ No. 128

    A spring-mass system has m = 4 kg and k = 400 N m⁻¹. The angular frequency is:

    a) 5 rad s⁻¹

    b) 10 rad s⁻¹

    c) 15 rad s⁻¹

    d) 20 rad s⁻¹

    Correct Answer: b) 10 rad s⁻¹

    Explanation:

    ω=4004=100=10 rad s−1\omega=\sqrt{\frac{400}{4}} =\sqrt{100}=10\text{ rad s}^{-1}

    MCQ No. 129

    If the mass attached to a spring is doubled while the spring constant remains unchanged, the time period becomes:

    a) Half

    b) √2 times

    c) Double

    d) Four times

    Correct Answer: b) √2 times

    Explanation: Since

    T∝mT\propto\sqrt{m}

    doubling the mass increases the time period by √2.

    MCQ No. 130

    If the spring constant is increased four times while the mass remains constant, the time period becomes:

    a) Four times

    b) Double

    c) Half

    d) One-fourth

    Correct Answer: c) Half

    Explanation: Since

    T∝1kT\propto\frac1{\sqrt{k}}

    quadrupling k halves the time period.

    MCQ No. 131

    A spring stretches 8 cm under a force of 16 N. The spring constant is:

    a) 100 N m⁻¹

    b) 150 N m⁻¹

    c) 200 N m⁻¹

    d) 250 N m⁻¹

    Correct Answer: c) 200 N m⁻¹

    Explanation:

    k=160.08=200 N m−1k=\frac{16}{0.08}=200\text{ N m}^{-1}

    MCQ No. 132

    A spring with k = 500 N m⁻¹ is stretched by 0.04 m. The restoring force is:

    a) 10 N

    b) 15 N

    c) 20 N

    d) 25 N

    Correct Answer: c) 20 N

    Explanation:

    F=kx=500(0.04)=20 NF=kx=500(0.04)=20\text{ N}

    MCQ No. 133

    A simple pendulum has a length of 1 m. Taking g = 9.8 m s⁻², its time period is approximately:

    a) 1.0 s

    b) 1.5 s

    c) 2.0 s

    d) 2.5 s

    Correct Answer: c) 2.0 s

    Explanation:

    T=2π19.8≈2.0 sT=2\pi\sqrt{\frac{1}{9.8}} \approx2.0\text{ s}

    MCQ No. 134

    The length of a pendulum is increased from 1 m to 4 m. Its time period becomes:

    a) Half

    b) Double

    c) Four times

    d) Unchanged

    Correct Answer: b) Double

    Explanation: Since

    T∝LT\propto\sqrt{L}
    4=2\sqrt4=2

    MCQ No. 135

    If the length of a pendulum becomes one-fourth of its original value, the time period becomes:

    a) One-fourth

    b) Half

    c) Double

    d) Four times

    Correct Answer: b) Half

    Explanation:

    T∝LT\propto\sqrt{L}
    14=12\sqrt{\frac14}=\frac12

    MCQ No. 136

    A pendulum has a time period of 2 s on Earth. Assuming g on the Moon is one-sixth of that on Earth, its time period on the Moon is approximately:

    a) 0.82 s

    b) 2 s

    c) 4.9 s

    d) 12 s

    Correct Answer: c) 4.9 s

    Explanation:

    T∝1gT\propto\frac1{\sqrt g}
    TM=26≈4.9 sT_M=2\sqrt6\approx4.9\text{ s}

    MCQ No. 137

    When a pendulum is taken to a place where g is greater, its frequency:

    a) Decreases

    b) Increases

    c) Remains constant

    d) Becomes zero

    Correct Answer: b) Increases

    Explanation: As g increases, the time period decreases and frequency increases.

    MCQ No. 138

    If a pendulum clock is taken from Earth to the Moon without adjustment, it will:

    a) Run faster

    b) Run slower

    c) Keep correct time

    d) Stop immediately

    Correct Answer: b) Run slower

    Explanation: The larger time period on the Moon causes the clock to lose time.

    MCQ No. 139

    A pendulum inside an accelerating lift moving upward will have:

    a) Smaller time period

    b) Larger time period

    c) Zero time period

    d) Infinite time period

    Correct Answer: a) Smaller time period

    Explanation: Effective gravity increases, reducing the time period.

    MCQ No. 140

    A pendulum inside a lift accelerating downward has:

    a) Smaller time period

    b) Larger time period

    c) Unchanged time period

    d) Zero frequency

    Correct Answer: b) Larger time period

    Explanation: Effective gravity decreases, increasing the time period.

    MCQ No. 141

    If the lift is in free fall, the pendulum will:

    a) Oscillate normally

    b) Oscillate faster

    c) Stop oscillating

    d) Have double frequency

    Correct Answer: c) Stop oscillating

    Explanation: During free fall, effective gravity becomes zero, so no restoring force acts.

    MCQ No. 142

    Two identical springs connected in series have an equivalent spring constant equal to:

    a) k

    b) 2k

    c) k/2

    d) 4k

    Correct Answer: c) k/2

    Explanation:

    keq=k2k_{eq}=\frac{k}{2}

    MCQ No. 143

    Two identical springs connected in parallel have an equivalent spring constant:

    a) k/2

    b) k

    c) 2k

    d) 4k

    Correct Answer: c) 2k

    Explanation:

    keq=2kk_{eq}=2k

    MCQ No. 144

    Compared with a single spring, the time period using two identical springs in parallel is:

    a) Larger

    b) √2 times larger

    c) Smaller

    d) Double

    Correct Answer: c) Smaller

    Explanation: A larger equivalent spring constant decreases the time period.

    MCQ No. 145

    Compared with a single spring, the time period using two identical springs in series is:

    a) Smaller

    b) Larger

    c) Unchanged

    d) Zero

    Correct Answer: b) Larger

    Explanation: A smaller equivalent spring constant increases the time period.

    MCQ No. 146

    A spring stores 18 J of elastic potential energy. If its extension is doubled, the stored energy becomes:

    a) 18 J

    b) 36 J

    c) 54 J

    d) 72 J

    Correct Answer: d) 72 J

    Explanation:

    Elastic potential energy is

    U=12kx2U=\frac12kx^2

    Doubling x makes the energy four times.

    MCQ No. 147

    A particle in SHM has an amplitude of 0.20 m. Its maximum displacement from the equilibrium position is:

    a) 0.10 m

    b) 0.20 m

    c) 0.30 m

    d) 0.40 m

    Correct Answer: b) 0.20 m

    Explanation: The amplitude is the maximum displacement.

    MCQ No. 148

    A spring requires 40 N to produce an extension of 0.20 m. What force is required for an extension of 0.30 m?

    a) 40 N

    b) 50 N

    c) 60 N

    d) 80 N

    Correct Answer: c) 60 N

    Explanation:

    k=400.20=200 N m−1k=\frac{40}{0.20}=200\text{ N m}^{-1}
    F=kx=200(0.30)=60 NF=kx=200(0.30)=60\text{ N}

    MCQ No. 149

    The time period of a spring-mass system is 4 s. If the attached mass becomes four times greater, the new time period will be:

    a) 2 s

    b) 4 s

    c) 8 s

    d) 16 s

    Correct Answer: c) 8 s

    Explanation:

    T∝mT\propto\sqrt m

    Increasing the mass four times doubles the time period.

    MCQ No. 150

    The spring constant of a spring is increased from 200 N m⁻¹ to 800 N m⁻¹. If the mass remains unchanged, the angular frequency becomes:

    a) Half

    b) Double

    c) Four times

    d) Unchanged

    Correct Answer: b) Double

    Explanation:

    Since

    ω=km\omega=\sqrt{\frac{k}{m}}

    quadrupling the spring constant doubles the angular frequency.

    MCQ No. 151

    A particle executes SHM with an amplitude of 0.20 m and an angular frequency of 10 rad s⁻¹. Its maximum velocity is:

    a) 1 m s⁻¹

    b) 2 m s⁻¹

    c) 4 m s⁻¹

    d) 20 m s⁻¹

    Correct Answer: b) 2 m s⁻¹

    Explanation:

    vmax⁡=Aω=(0.20)(10)=2 m s−1v_{\max}=A\omega=(0.20)(10)=2\text{ m s}^{-1}

    MCQ No. 152

    A particle has an amplitude of 0.10 m and an angular frequency of 20 rad s⁻¹. The maximum acceleration is:

    a) 20 m s⁻²

    b) 30 m s⁻²

    c) 40 m s⁻²

    d) 80 m s⁻²

    Correct Answer: c) 40 m s⁻²

    Explanation:

    amax⁡=Aω2=(0.10)(20)2=40 m s−2a_{\max}=A\omega^2 =(0.10)(20)^2 =40\text{ m s}^{-2}

    MCQ No. 153

    A particle has an amplitude of 5 cm and an angular frequency of 40 rad s⁻¹. The maximum velocity is:

    a) 1 m s⁻¹

    b) 2 m s⁻¹

    c) 3 m s⁻¹

    d) 4 m s⁻¹

    Correct Answer: b) 2 m s⁻¹

    Explanation:

    A=0.05 mA=0.05\text{ m}
    vmax⁡=Aω=(0.05)(40)=2 m s−1v_{\max}=A\omega=(0.05)(40)=2\text{ m s}^{-1}

    MCQ No. 154

    The acceleration of a particle executing SHM is −16 m s⁻² when its displacement is 0.04 m. The value of ω\omega is:

    a) 10 rad s⁻¹

    b) 20 rad s⁻¹

    c) 30 rad s⁻¹

    d) 40 rad s⁻¹

    Correct Answer: b) 20 rad s⁻¹

    Explanation:

    a=−ω2xa=-\omega^2x
    16=ω2(0.04)16=\omega^2(0.04)
    ω2=400\omega^2=400
    ω=20 rad s−1\omega=20\text{ rad s}^{-1}

    MCQ No. 155

    The maximum acceleration of an oscillator is 45 m s⁻². If its amplitude is 0.05 m, the angular frequency is:

    a) 20 rad s⁻¹

    b) 25 rad s⁻¹

    c) 30 rad s⁻¹

    d) 40 rad s⁻¹

    Correct Answer: c) 30 rad s⁻¹

    Explanation:

    45=0.05ω245=0.05\omega^2
    ω2=900\omega^2=900
    ω=30 rad s−1\omega=30\text{ rad s}^{-1}

    MCQ No. 156

    A particle has A = 0.20 m and ω = 5 rad s⁻¹. Its maximum acceleration is:

    a) 2.5 m s⁻²

    b) 5 m s⁻²

    c) 10 m s⁻²

    d) 20 m s⁻²

    Correct Answer: b) 5 m s⁻²

    Explanation:

    amax⁡=Aω2=(0.20)(25)=5 m s−2a_{\max}=A\omega^2 =(0.20)(25)=5\text{ m s}^{-2}

    MCQ No. 157

    A particle executes SHM with amplitude 0.25 m and angular frequency 8 rad s⁻¹. Its maximum speed is:

    a) 1 m s⁻¹

    b) 2 m s⁻¹

    c) 3 m s⁻¹

    d) 4 m s⁻¹

    Correct Answer: b) 2 m s⁻¹

    Explanation:

    vmax⁡=Aω=(0.25)(8)=2 m s−1v_{\max}=A\omega =(0.25)(8)=2\text{ m s}^{-1}

    MCQ No. 158

    If the amplitude of SHM is doubled while the angular frequency remains constant, the maximum velocity becomes:

    a) Half

    b) Double

    c) Four times

    d) Unchanged

    Correct Answer: b) Double

    Explanation:

    Since

    vmax⁡=Aωv_{\max}=A\omega

    doubling the amplitude doubles the maximum velocity.

    MCQ No. 159

    If the angular frequency is doubled while the amplitude remains constant, the maximum acceleration becomes:

    a) Double

    b) Four times

    c) Half

    d) Unchanged

    Correct Answer: b) Four times

    Explanation:

    amax⁡=Aω2a_{\max}=A\omega^2

    Doubling ω\omega increases amax⁡a_{\max}  by 4.

    MCQ No. 160

    A particle executes SHM with A = 0.10 m and ω = 50 rad s⁻¹. The maximum acceleration is:

    a) 50 m s⁻²

    b) 100 m s⁻²

    c) 250 m s⁻²

    d) 500 m s⁻²

    Correct Answer: c) 250 m s⁻²

    Explanation:

    amax⁡=Aω2=(0.10)(2500)=250a_{\max}=A\omega^2 =(0.10)(2500)=250

    MCQ No. 161

    The total mechanical energy of a spring oscillator is 18 J. At the mean position, its kinetic energy is:

    a) 0 J

    b) 6 J

    c) 9 J

    d) 18 J

    Correct Answer: d) 18 J

    Explanation: At the mean position, potential energy is zero, so all the mechanical energy is kinetic.

    MCQ No. 162

    The total energy of an oscillator is 24 J. At an extreme position, its potential energy is:

    a) 0 J

    b) 12 J

    c) 24 J

    d) 48 J

    Correct Answer: c) 24 J

    Explanation: At the extreme position, velocity is zero, so all the energy is potential.

    MCQ No. 163

    At the mean position of SHM, the potential energy is:

    a) Maximum

    b) Half of total energy

    c) Zero

    d) Equal to kinetic energy

    Correct Answer: c) Zero

    Explanation: The displacement is zero at the mean position; therefore, the potential energy is zero.

    MCQ No. 164

    At the extreme position, the kinetic energy of an oscillator is:

    a) Maximum

    b) Minimum (zero)

    c) Half of total energy

    d) Equal to potential energy

    Correct Answer: b) Minimum (zero)

    Explanation: The particle momentarily stops at the extreme position.

    MCQ No. 165

    If the amplitude of SHM becomes three times greater, the total mechanical energy becomes:

    a) Three times

    b) Six times

    c) Nine times

    d) Twelve times

    Correct Answer: c) Nine times

    Explanation:

    E∝A2E\propto A^2
    (3A)2=9A2(3A)^2=9A^2

    MCQ No. 166

    A spring stores 8 J of elastic potential energy. If the extension becomes three times greater, the stored energy becomes:

    a) 16 J

    b) 24 J

    c) 48 J

    d) 72 J

    Correct Answer: d) 72 J

    Explanation:

    U=12kx2U=\frac12kx^2

    Tripling the extension makes the energy 9 times greater.

    MCQ No. 167

    The elastic potential energy stored in a spring depends on:

    a) Extension only

    b) Spring constant only

    c) Both spring constant and square of extension

    d) Mass attached

    Correct Answer: c) Both spring constant and square of extension

    Explanation:

    U=12kx2U=\frac12kx^2

    MCQ No. 168

    A spring with k = 400 N m⁻¹ is stretched by 0.20 m. The elastic potential energy stored is:

    a) 2 J

    b) 4 J

    c) 8 J

    d) 16 J

    Correct Answer: c) 8 J

    Explanation:

    U=12(400)(0.20)2=8 JU=\frac12(400)(0.20)^2 =8\text{ J}

    MCQ No. 169

    A spring stores 18 J when stretched by 0.30 m. If stretched by 0.60 m, the energy becomes:

    a) 18 J

    b) 36 J

    c) 54 J

    d) 72 J

    Correct Answer: d) 72 J

    Explanation: Doubling the extension increases the energy fourfold.

    MCQ No. 170

    During SHM, the sum of kinetic and potential energies remains:

    a) Increasing

    b) Decreasing

    c) Constant

    d) Zero

    Correct Answer: c) Constant

    Explanation: In ideal SHM, total mechanical energy is conserved.

    MCQ No. 171

    The velocity of an oscillator is maximum because:

    a) Potential energy is maximum

    b) Restoring force is maximum

    c) Potential energy is minimum

    d) Acceleration is maximum

    Correct Answer: c) Potential energy is minimum

    Explanation: At the mean position, potential energy is minimum (zero), so kinetic energy and velocity are maximum.

    MCQ No. 172

    The acceleration is greatest where:

    a) Velocity is maximum

    b) Displacement is zero

    c) Displacement is maximum

    d) Kinetic energy is maximum

    Correct Answer: c) Displacement is maximum

    Explanation: Since a=−ω2xa=-\omega^2x, acceleration is greatest at the extreme positions.

    MCQ No. 173

    The ratio of maximum kinetic energy to total mechanical energy in SHM is:

    a) 0

    b) 1/2

    c) 1

    d) 2

    Correct Answer: c) 1

    Explanation: At the mean position, kinetic energy equals the total mechanical energy.

    MCQ No. 174

    If the amplitude is halved, the total mechanical energy becomes:

    a) Half

    b) One-fourth

    c) Double

    d) Four times

    Correct Answer: b) One-fourth

    Explanation:

    E∝A2E\propto A^2 (12)2=14\left(\frac12\right)^2=\frac14

    MCQ No. 175

    A spring oscillator stores 50 J of total mechanical energy. At the mean position, its potential energy is:

    a) 0 J

    b) 25 J

    c) 50 J

    d) 100 J

    Correct Answer: a) 0 J

    Explanation: At the equilibrium position, displacement is zero, so the elastic potential energy is zero and the entire 50 J is kinetic energy.

    MCQ No. 176

    A spring of force constant 400 N m⁻¹ is stretched by 0.10 m. The elastic potential energy stored is:

    a) 1 J

    b) 2 J

    c) 4 J

    d) 8 J

    Correct Answer: b) 2 J

    Explanation:

    U=12kx2=12(400)(0.10)2=2 JU=\frac{1}{2}kx^2=\frac{1}{2}(400)(0.10)^2=2\text{ J}

    MCQ No. 177

    A particle performs SHM with amplitude 0.25 m and angular frequency 12 rad s⁻¹. Its maximum acceleration is:

    a) 24 m s⁻²

    b) 30 m s⁻²

    c) 36 m s⁻²

    d) 40 m s⁻²

    Correct Answer: c) 36 m s⁻²

    Explanation:

    amax=Aω2=(0.25)(12)2=36 m s−2a_{max}=A\omega^2=(0.25)(12)^2=36\text{ m s}^{-2}

    MCQ No. 178

    A spring-mass system has m = 1 kg and k = 400 N m⁻¹. The time period is:

    a) 0.157 s

    b) 0.314 s

    c) 0.628 s

    d) 1.257 s

    Correct Answer: b) 0.314 s

    Explanation:

    T=2πmk=2π1400=2π20=0.314 sT=2\pi\sqrt{\frac{m}{k}} =2\pi\sqrt{\frac{1}{400}} =\frac{2\pi}{20} =0.314\text{ s}

    MCQ No. 179

    The natural frequency of a spring-mass system is 5 Hz. If an external force of 5 Hz acts on it, the system undergoes:

    a) Damping

    b) Resonance

    c) Free oscillations

    d) Critical damping

    Correct Answer: b) Resonance

    Explanation: Resonance occurs when the driving frequency equals the natural frequency.

    MCQ No. 180

    The amplitude of forced oscillations becomes maximum when:

    a) Driving frequency is zero

    b) Driving frequency is less than the natural frequency

    c) Driving frequency equals the natural frequency

    d) Driving frequency is greater than the natural frequency

    Correct Answer: c) Driving frequency equals the natural frequency

    Explanation: This condition produces resonance.

    MCQ No. 181

    A pendulum clock is taken from sea level to the top of a mountain. It will:

    a) Gain time

    b) Lose time

    c) Keep correct time

    d) Stop immediately

    Correct Answer: b) Lose time

    Explanation: Gravity decreases at higher altitudes, increasing the time period and causing the clock to lose time.

    MCQ No. 182

    If the effective value of g decreases by 19%, the time period of a pendulum will:

    a) Decrease

    b) Increase

    c) Remain unchanged

    d) Become zero

    Correct Answer: b) Increase

    Explanation: Since

    T∝1gT\propto\frac1{\sqrt g}

    a decrease in gravity increases the time period.

    MCQ No. 183

    A pendulum clock is taken deep into a mine. Compared with the Earth's surface, it will:

    a) Run faster

    b) Run slower

    c) Show no change

    d) Stop permanently

    Correct Answer: b) Run slower

    Explanation: Gravity decreases inside the Earth, increasing the time period.

    MCQ No. 184

    Two identical springs each having spring constant k are connected in series. Compared with a single spring, the angular frequency becomes:

    a) √2ω

    b) ω/√2

    c) 2ω

    d) ω

    Correct Answer: b) ω/√2

    Explanation:

    keq=k2k_{eq}=\frac{k}{2}
    ω=km\omega=\sqrt{\frac{k}{m}}

    Hence,

    ω′=ω2\omega'=\frac{\omega}{\sqrt2}

    MCQ No. 185

    Two identical springs are connected in parallel. The angular frequency becomes:

    a) ω/2

    b) ω

    c) √2ω

    d) 2ω

    Correct Answer: c) √2ω

    Explanation:

    For parallel combination,

    keq=2kk_{eq}=2k

    Therefore,

    ω′=2 ω\omega'=\sqrt2\,\omega

    MCQ No. 186

    The energy supplied to a resonating system is mainly used to:

    a) Increase its mass

    b) Increase its amplitude

    c) Reduce its frequency

    d) Change its spring constant

    Correct Answer: b) Increase its amplitude

    Explanation: At resonance, maximum energy transfer produces maximum amplitude.

    MCQ No. 187

    Which one of the following systems should be critically damped?

    a) Pendulum clock

    b) Tuning fork

    c) Car suspension

    d) Guitar string

    Correct Answer: c) Car suspension

    Explanation: Critical damping enables the suspension to return to equilibrium quickly without oscillating.

    MCQ No. 188

    A particle executes SHM with A = 0.40 m and ω = 15 rad s⁻¹. The maximum velocity is:

    a) 3 m s⁻¹

    b) 4.5 m s⁻¹

    c) 6 m s⁻¹

    d) 9 m s⁻¹

    Correct Answer: c) 6 m s⁻¹

    Explanation:

    vmax=Aω=(0.40)(15)=6 m s−1v_{max}=A\omega=(0.40)(15)=6\text{ m s}^{-1}

    MCQ No. 189

    A spring stores 50 J of energy. If the extension is reduced to half, the stored energy becomes:

    a) 12.5 J

    b) 25 J

    c) 50 J

    d) 100 J

    Correct Answer: a) 12.5 J

    Explanation:

    Since

    U∝x2U\propto x^2

    Halving the extension reduces the energy to one-fourth.

    MCQ No. 190

    A particle executes SHM with amplitude 0.20 m. At a displacement of 0.10 m, the restoring force is:

    a) Zero

    b) Half of the maximum restoring force

    c) Equal to the maximum restoring force

    d) Double the maximum restoring force

    Correct Answer: b) Half of the maximum restoring force

    Explanation: Since

    F=−kxF=-kx

    the restoring force is directly proportional to displacement.

    MCQ No. 191

    The quality of a resonant system is improved by:

    a) Increasing damping

    b) Eliminating restoring force

    c) Reducing damping

    d) Increasing friction

    Correct Answer: c) Reducing damping

    Explanation: Lower damping results in sharper and stronger resonance.

    MCQ No. 192

    The frequency of free oscillations depends mainly upon:

    a) Amplitude

    b) External force

    c) Physical properties of the system

    d) Air pressure only

    Correct Answer: c) Physical properties of the system

    Explanation: Mass, elasticity, and dimensions determine the natural frequency.

    MCQ No. 193

    A spring oscillator has m = 2 kg and k = 800 N m⁻¹. Its angular frequency is:

    a) 10 rad s⁻¹

    b) 20 rad s⁻¹

    c) 30 rad s⁻¹

    d) 40 rad s⁻¹

    Correct Answer: b) 20 rad s⁻¹

    Explanation:

    ω=8002=400=20 rad s−1\omega=\sqrt{\frac{800}{2}} =\sqrt{400} =20\text{ rad s}^{-1}

    MCQ No. 194

    A pendulum has a length of 0.25 m. Taking g = 9.8 m s⁻², its approximate time period is:

    a) 0.5 s

    b) 1.0 s

    c) 1.5 s

    d) 2.0 s

    Correct Answer: b) 1.0 s

    Explanation:

    T=2π0.259.8≈1.0 sT=2\pi\sqrt{\frac{0.25}{9.8}} \approx1.0\text{ s}

    MCQ No. 195

    The motion of a sewing machine needle is approximately:

    a) Projectile motion

    b) Uniform circular motion

    c) Oscillatory motion

    d) Random motion

    Correct Answer: c) Oscillatory motion

    Explanation: The needle moves repeatedly to and fro, making its motion oscillatory.

    MCQ No. 196

    A body performing SHM has its maximum kinetic energy when its displacement is:

    a) Maximum

    b) Half the amplitude

    c) Zero

    d) Equal to the amplitude

    Correct Answer: c) Zero

    Explanation: At the equilibrium position, the velocity and kinetic energy are maximum.

    MCQ No. 197

    The restoring force becomes maximum when the particle is at:

    a) Mean position

    b) Quarter amplitude

    c) Half amplitude

    d) Extreme position

    Correct Answer: d) Extreme position

    Explanation: Since restoring force is proportional to displacement, it is greatest at the extreme positions.

    MCQ No. 198

    The total mechanical energy of SHM depends directly upon:

    a) Frequency

    b) Time period

    c) Square of amplitude

    d) Mass only

    Correct Answer: c) Square of amplitude

    Explanation: The total energy of SHM is proportional to A2A^2.

    MCQ No. 199

    Which one of the following statements about resonance is correct?

    a) It occurs only in pendulums.

    b) It occurs when damping is maximum.

    c) It occurs when the driving frequency equals the natural frequency.

    d) It decreases the amplitude of oscillation.

    Correct Answer: c) It occurs when the driving frequency equals the natural frequency.

    Explanation: Resonance is the condition of maximum amplitude due to matching driving and natural frequencies.

    MCQ No. 200

    A spring-mass system and a simple pendulum both execute SHM. The time period of the spring-mass system depends upon:

    a) Length and gravity

    b) Mass and spring constant

    c) Mass and amplitude only

    d) Gravity and amplitude

    Correct Answer: b) Mass and spring constant

    Explanation: The time period of a spring-mass system is given by

    T=2πmkT=2\pi\sqrt{\frac{m}{k}}

    It depends only on the attached mass and the spring constant (for ideal SHM), unlike the simple pendulum, whose period depends on length and gravitational acceleration.

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