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7. The pendulum in a grandfather clock is made of brass and keeps perfect time at `17°C`. How much time is gained or lost in a year if the clock is kept at `25°C`? (Assume the frequency dependence on length for a simple pendulum applies.)
7. The pendulum in a grandfather clock is made of brass and keeps perfect time at 17°C. How much time is gained or lost in a year if the clock is kept at 25°C? (Assume the frequency dependence on length for a simple pendulum applies.) about 40 mins
Given:
`T_0 = 17°C`
`T = 25°C`
Equation:
period `t = 2πsqrt(L/g)`
`∆L = αL_0∆T `
Solution:
the fractional lost or gained will be `∆t/t_0`
`(∆t)/t_0 = (t-t_0)/t_0 = (2πsqrt(L/g)-2πsqrt(L_0/g))/(2πsqrt(L_0/g))`
`(∆t)/t_0 =(sqrt(L)-sqrt(L_0))/sqrt(L_0)`
`(∆t)/t_0 =(sqrt(L_0+∆L)-sqrt(L_0))/sqrt(L_0)`
`(∆t)/t_0 =(sqrt(L_0+ αL_0∆T )-sqrt(L_0))/sqrt(L_0)`
`(∆t)/t_0 =sqrt(1+ α∆T )-1`
`(∆t)/t_0 =sqrt(1+ (19times10^(-6) //C°)(25°C-17°C) )-1`
`(∆t)/t_0 =7.60 times 10^(-5)`
For one year, we have
`t = 7.60 times 10^(-5) (3.16 times 10^7 s) = 2402\ s≈ 40\ min`
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