Pendulum Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Pendulum Motion MCQs with step-by-step solutions (33 questions). Part of Oscillations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Pendulum Motion · easy · theory
The period of a simple pendulum of length $L$ at a place where the acceleration due to gravity is $g$ is:
A. $2\pi\sqrt{\dfrac{L}{g}}$ ✓ Correct
B. $\dfrac{1}{2\pi}\sqrt{\dfrac{L}{g}}$
C. $2\pi\dfrac{L}{g}$
D. $2\pi\sqrt{\dfrac{g}{L}}$
Solution: For small oscillations the restoring force gives $\omega = \sqrt{\dfrac{g}{L}}$, so $T = 2\pi\sqrt{\dfrac{L}{g}}$.
Q2 — Pendulum Motion · easy · theory
The period of a simple pendulum is independent of:
A. All of these
B. The length of the pendulum
C. The mass of the bob ✓ Correct
D. The acceleration due to gravity
Solution: The mass cancels from the equation of motion, so $T = 2\pi\sqrt{\dfrac{L}{g}}$ contains only the length and $g$.
Q3 — Pendulum Motion · easy · numerical
If the length of a simple pendulum is made four times its original value, its period becomes:
A. Unchanged
B. Twice as long ✓ Correct
C. Four times as long
D. Half as long
Solution: $T \propto \sqrt{L}$, so quadrupling the length doubles the period.
Q4 — Pendulum Motion · hard · numerical
A simple pendulum suspended from the ceiling of a lift has period $T$. When the lift accelerates upward with acceleration $\dfrac{g}{3}$, the new period is:
A. $\dfrac{2}{\sqrt{3}}T$
B. $\dfrac{4}{3}T$
C. $\dfrac{\sqrt{3}}{2}T$ ✓ Correct
D. $\dfrac{3}{4}T$
Solution: The effective gravity becomes $g_{eff} = g + \dfrac{g}{3} = \dfrac{4g}{3}$. Hence $T' = 2\pi\sqrt{\dfrac{L}{4g/3}} = \sqrt{\dfrac{3}{4}}\,T = \dfrac{\sqrt{3}}{2}T$.
Q5 — Pendulum Motion · medium · theory
The period of a simple pendulum inside a freely falling lift is:
A. Infinite ✓ Correct
B. Unchanged
C. Zero
D. Halved
Solution: In free fall $g_{eff} = 0$, so $T = 2\pi\sqrt{\dfrac{L}{0}} \to \infty$. The bob simply does not oscillate — it floats with the lift.
Q6 — Pendulum Motion · medium · theory
The period of a simple pendulum inside a satellite orbiting freely around the Earth is:
A. The same as on Earth
B. Two seconds
C. Infinite ✓ Correct
D. Zero
Solution: A freely orbiting satellite is in continuous free fall, so the effective gravity inside is zero and the pendulum does not oscillate at all.
Q7 — Pendulum Motion · medium · theory
A simple pendulum in a lift accelerating downward with acceleration $a$ (where $a < g$) has an effective gravitational acceleration of:
A. $\dfrac{g}{a}$
B. $g - a$ ✓ Correct
C. $\sqrt{g^2 + a^2}$
D. $g + a$
Solution: In the accelerating frame a pseudo-force acts upward on the bob, reducing the net downward pull to $m(g - a)$, so the pendulum swings more slowly.
Q8 — Pendulum Motion · easy · theory
A seconds pendulum is one whose period is:
A. $1\text{ s}$
B. $0.5\text{ s}$
C. $4\text{ s}$
D. $2\text{ s}$ ✓ Correct
Solution: A seconds pendulum takes one second for each swing from one extreme to the other, so a complete oscillation takes $2\text{ s}$. Its length on Earth is about $0.993\text{ m}$.
Q9 — Pendulum Motion · medium · numerical
The period of a simple pendulum of length $1\text{ m}$ at a place where $g = 9.8\text{ m/s}^2$ is approximately:
A. $3.14\text{ s}$
B. $6.28\text{ s}$
C. $2.01\text{ s}$ ✓ Correct
D. $1.00\text{ s}$
Solution: $T = 2\pi\sqrt{\dfrac{1}{9.8}} = 2\pi(0.3194) \approx 2.01\text{ s}$.
Q10 — Pendulum Motion · medium · numerical
A simple pendulum is taken to the Moon, where the acceleration due to gravity is one-sixth of that on the Earth. Its period becomes:
A. $\dfrac{1}{\sqrt{6}}$ times as large
B. Unchanged
C. $6$ times as large
D. $\sqrt{6}$ times as large ✓ Correct
Solution: $T \propto \dfrac{1}{\sqrt{g}}$, so reducing $g$ to $\dfrac{g}{6}$ multiplies the period by $\sqrt{6} \approx 2.45$.
Q11 — Pendulum Motion · medium · theory
The oscillation of a simple pendulum is simple harmonic only when:
A. The bob is very heavy
B. The angular amplitude is small, so that $\sin\theta \approx \theta$ ✓ Correct
C. The string is extensible
D. The angular amplitude is large
Solution: The restoring torque is proportional to $\sin\theta$, not $\theta$. Only for small angles does $\sin\theta \approx \theta$ make the motion truly simple harmonic.
Q12 — Pendulum Motion · easy · numerical
If the acceleration due to gravity at a place is doubled, the period of a given simple pendulum becomes:
A. $\dfrac{1}{\sqrt{2}}$ times its original value ✓ Correct
B. Twice its original value
C. $\sqrt{2}$ times its original value
D. Half its original value
Solution: $T \propto \dfrac{1}{\sqrt{g}}$, so doubling $g$ divides the period by $\sqrt{2}$.
Q13 — Pendulum Motion · easy · theory
The effective length of a simple pendulum is measured from the point of suspension to the:
A. Centre of gravity of the bob ✓ Correct
B. Top of the bob
C. Midpoint of the string only
D. Bottom of the bob
Solution: The bob behaves as a point mass located at its centre of gravity, so the effective length is the string length plus the radius of the bob.
Q14 — Pendulum Motion · hard · theory
A pendulum clock that keeps correct time at sea level is taken to the top of a high mountain. It will:
A. Gain time, because $g$ decreases
B. Lose time, because $g$ decreases and the period increases ✓ Correct
C. Keep correct time
D. Stop oscillating altogether
Solution: Gravity is weaker at altitude, so $T = 2\pi\sqrt{\dfrac{L}{g}}$ grows. Each swing takes longer, and the clock falls behind.
Q15 — Pendulum Motion · medium · theory
The period of a simple pendulum of length $L$ in a lift moving upward with constant velocity is:
A. $2\pi\sqrt{\dfrac{L}{g}}$, the same as when at rest ✓ Correct
B. Zero
C. $2\pi\sqrt{\dfrac{L}{2g}}$
D. Infinite
Solution: Constant velocity means zero acceleration, so the lift is an inertial frame and the effective gravity is unchanged at $g$.
Q16 — Pendulum Motion · medium · theory
For a simple pendulum, a graph of $T^2$ against $L$ is:
A. A straight line through the origin with slope $\dfrac{4\pi^2}{g}$ ✓ Correct
B. A parabola
C. A hyperbola
D. A straight line with a negative slope
Solution: Squaring $T = 2\pi\sqrt{\dfrac{L}{g}}$ gives $T^2 = \dfrac{4\pi^2}{g}L$, a linear relation whose slope is used experimentally to determine $g$.
Q17 — Pendulum Motion · hard · theory
A simple pendulum oscillates with angular amplitude $\theta_0$. Its speed at the lowest point of the swing, for a length $L$, is:
A. $\sqrt{gL\sin\theta_0}$
B. $\sqrt{2gL(1 - \cos\theta_0)}$ ✓ Correct
C. $\sqrt{2gL\cos\theta_0}$
D. $\sqrt{2gL}$
Solution: The bob falls a vertical height $h = L(1 - \cos\theta_0)$ from the extreme to the lowest point, so energy conservation gives $v = \sqrt{2gh}$.
Q18 — Pendulum Motion · medium · numerical
The period of a simple pendulum of length $0.4\text{ m}$ is ($g = 10\text{ m/s}^2$):
A. $2.51\text{ s}$
B. $0.20\text{ s}$
C. $0.63\text{ s}$
D. $1.26\text{ s}$ ✓ Correct
Solution: $T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{0.4}{10}} = 2\pi\sqrt{0.04} = 2\pi(0.2) \approx 1.26\text{ s}$.
Q19 — Pendulum Motion · hard · numerical
The length of a seconds pendulum at a place where $g = 9.8\text{ m/s}^2$ is approximately:
A. $1.986\text{ m}$
B. $0.248\text{ m}$
C. $0.993\text{ m}$ ✓ Correct
D. $2.00\text{ m}$
Solution: A seconds pendulum has $T = 2\text{ s}$, so $L = \dfrac{gT^2}{4\pi^2} = \dfrac{9.8 \times 4}{39.48} \approx 0.993\text{ m}$.
Q20 — Pendulum Motion · easy · numerical
A simple pendulum has a period of $2\text{ s}$. If its length is made four times as long, its new period is:
A. $2\text{ s}$
B. $8\text{ s}$
C. $1\text{ s}$
D. $4\text{ s}$ ✓ Correct
Solution: $T \propto \sqrt{L}$, so quadrupling the length doubles the period to $4\text{ s}$.
Q21 — Pendulum Motion · hard · numerical
A simple pendulum of period $T$ is placed in a lift accelerating upward with acceleration $g$. Its new period is:
A. $0.5T$
B. $1.414T$
C. $0.707T$ ✓ Correct
D. $2T$
Solution: The effective gravity becomes $g_{eff} = g + g = 2g$, so $T' = \dfrac{T}{\sqrt{2}} \approx 0.707T$.
Q22 — Pendulum Motion · medium · numerical
The period of a simple pendulum of length $0.9\text{ m}$ is ($g = 10\text{ m/s}^2$):
A. $0.94\text{ s}$
B. $3.77\text{ s}$
C. $0.30\text{ s}$
D. $1.88\text{ s}$ ✓ Correct
Solution: $T = 2\pi\sqrt{\dfrac{0.9}{10}} = 2\pi\sqrt{0.09} = 2\pi(0.3) \approx 1.88\text{ s}$.
Q23 — Pendulum Motion · medium · numerical
The period of a simple pendulum of length $2.5\text{ m}$ is ($g = 10\text{ m/s}^2$):
A. $0.50\text{ s}$
B. $6.28\text{ s}$
C. $3.14\text{ s}$ ✓ Correct
D. $1.57\text{ s}$
Solution: $T = 2\pi\sqrt{\dfrac{2.5}{10}} = 2\pi\sqrt{0.25} = 2\pi(0.5) \approx 3.14\text{ s}$.
Q24 — Pendulum Motion · easy · numerical
A simple pendulum of period $2\text{ s}$ has its length increased nine times. Its new period is:
A. $3\text{ s}$
B. $18\text{ s}$
C. $4.5\text{ s}$
D. $6\text{ s}$ ✓ Correct
Solution: $T \propto \sqrt{L}$, so a nine-fold length gives $\sqrt{9} = 3$ times the period: $T = 6\text{ s}$.
Q25 — Pendulum Motion · hard · numerical
A simple pendulum of period $T$ is in a lift accelerating downward with acceleration $\dfrac{g}{2}$. Its new period is:
A. $1.414T$ ✓ Correct
B. $0.5T$
C. $2T$
D. $0.707T$
Solution: The effective gravity is $g - \dfrac{g}{2} = \dfrac{g}{2}$, so $T' = T\sqrt{\dfrac{g}{g/2}} = T\sqrt{2} \approx 1.414T$.
Q26 — Pendulum Motion · hard · numerical
A pendulum with a period of $2\text{ s}$ on Earth is taken to the Moon, where gravity is one-sixth as strong. Its period there is approximately:
A. $12\text{ s}$
B. $4.9\text{ s}$ ✓ Correct
C. $2.0\text{ s}$
D. $0.82\text{ s}$
Solution: $T \propto \dfrac{1}{\sqrt{g}}$, so $T_{moon} = 2\sqrt{6} \approx 4.9\text{ s}$.
Q27 — Pendulum Motion · medium · numerical
The period of a simple pendulum of length $1.6\text{ m}$ is ($g = 10\text{ m/s}^2$):
A. $5.03\text{ s}$
B. $1.26\text{ s}$
C. $2.51\text{ s}$ ✓ Correct
D. $0.40\text{ s}$
Solution: $T = 2\pi\sqrt{0.16} = 2\pi(0.4) \approx 2.51\text{ s}$.
Q28 — Pendulum Motion · easy · numerical
The length of a simple pendulum is made twenty-five times as long. Its period becomes:
A. $5$ times as long ✓ Correct
B. $25$ times as long
C. $\dfrac{1}{5}$ as long
D. $2.5$ times as long
Solution: $T \propto \sqrt{L}$, so the period grows by $\sqrt{25} = 5$ times.
Q29 — Pendulum Motion · easy · numerical
If the acceleration due to gravity becomes four times as large, the period of a given simple pendulum becomes:
A. Four times as long
B. Half as long ✓ Correct
C. Twice as long
D. Unchanged
Solution: $T \propto \dfrac{1}{\sqrt{g}}$, so a four-fold $g$ halves the period.
Q30 — Pendulum Motion · hard · numerical
A simple pendulum of period $T$ is in a lift accelerating upward with acceleration $\dfrac{g}{4}$. Its new period is approximately:
A. $0.894T$ ✓ Correct
B. $0.8T$
C. $1.25T$
D. $1.118T$
Solution: The effective gravity is $\dfrac{5g}{4}$, so $T' = T\sqrt{\dfrac{4}{5}} \approx 0.894T$.