When the length of a simple pendulum is doubled, the ratio of the new frequency to the old frequency is:
Correct answer: D. 1/√2
- A. 1/4
- B. 1/2
- C. √2
- D. 1/√2
Explanation
f = 1/2π × √(g/L)where 'g' is the acceleration due to gravity and 'L' is the length of the pendulum.When the length of the pendulum is doubled, the new length becomes 2L. Therefore, the new frequency can be calculated using the same formula:f' = 1/2π × √(g/2L)To find the ratio of the new frequency to the old frequency, we can divide f' by f:f'/f = [1/2π × √(g/2L)] / [1/2π × √(g/L)]f'/f = √(L/2L)f'/f = √(1/2)f'/f = 1/√2Therefore, the correct option is D) 1/√2
This question appeared on the UHS MDCAT 2012 paper, which you can sit online with every answer explained.
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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