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A ball suspended from a thread has a frequency 2 Hz. When it is taken on planet X, the...
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The time period of a pendulum is closely approximated by the formula `T = 2*pi*sqrt(L/g)` where L is the length of the pendulum and g is the acceleration due to gravity.
On Earth, the frequency of the pendulum is 2 Hz. The time period is 1/2 = 0.5 seconds. `0.5 = 2*pi*sqrt(L/g_E)` . On planet X, the frequency of the same pendulum changes to 8 Hz. The time period of the pendulum is now 1/8 = 0.125 s. If the gravitational acceleration on planet X is `g_X` , `0.125 = 2*pi*sqrt(L/g_X)`
Divide the equations `0.5 = 2*pi*sqrt(L/g_E)` and `0.125 = 2*pi*sqrt(L/g_X)` . This gives: `4 = sqrt((L/g_E)/(L/g_X))`
=> `16 = g_X/g_E`
=> `g_X = 16*g_E`
The acceleration due to gravity on planet X is 16 times that of the gravitational attraction on Earth. The magnitude of the same is 156.8 m/s^2.
Posted by justaguide on September 23, 2013 at 1:13 PM (Answer #1)
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