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Question
The length of the second’s pendulum in a clock is increased to 4 times its initial length. Calculate the number of oscillations completed by the new pendulum in one minute.
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Solution
Time period of second's pendulum = 2 seconds
let the length of the pendulum by L
So, T = `2pi sqrt("L"/9)`
so, 2 = `2pi sqrt("L"/9)`
Now ,length becomes 4 times.
so,new time period be T
so, T = `2pi sqrt("4L"/9)`
`= (2pi sqrt("L"/9)) xx 2`
= T × 2 = 2 × 2 = 4 seconds
So,one oscillation is completed in 4 second
So, for 60 seconds or 1 minutes number of oscillations will be `= 1/4 xx 60 = 15`
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