Simple Pendulum - Derivation of Expression for its Time Period NEET Questions

Physics - Oscillations and Waves

Simple Pendulum - Derivation of Expression for its Time Period MCQ Questions


13.
The time period of a simple pendulum for small oscillations is:
A.
T = 2π√(g/L)
B.
T = π√(L/g)
C.
T = (1/2π)√(L/g)
D.
T = 2π√(L/g)
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ANSWER :
D. T = 2π√(L/g)
14.
In the formula T = 2π√(L/g), the time period T depends on:
A.
Only L (length) and g (gravitational acceleration)
B.
L, g, and m (mass of bob)
C.
Only g
D.
L, g, and A (amplitude of swing)
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ANSWER :
A. Only L (length) and g (gravitational acceleration)
15.
A pendulum of length L has time period T. If the length is increased to 4L (same location), the new time period is:
A.
2T
B.
T/2
C.
4T
D.
T/4
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ANSWER :
A. 2T
16.
If the length of a simple pendulum is halved, the time period becomes:
A.
T/√2
B.
2T
C.
T/2
D.
T√2
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ANSWER :
A. T/√2
17.
A simple pendulum has time period T₁ at a place where g = g₁. At another place where g = 4g₁ (same pendulum length), the time period T₂ is:
A.
2T₁
B.
4T₁
C.
T₁/2
D.
T₁/4
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ANSWER :
C. T₁/2
18.
The time period of a pendulum on the Moon (g_moon = 1.7 m/s²) compared to Earth (g_earth = 9.8 m/s²) for the same length L is approximately:
A.
5.76 times longer on Moon
B.
2.4 times longer on Moon
C.
2.4 times shorter on Moon
D.
Same on both
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ANSWER :
B. 2.4 times longer on Moon