Part A What is the change T in the period of a simple pendulum when the accelera
ID: 1481576 • Letter: P
Question
Part A
What is the change T in the period of a simple pendulum when the acceleration of gravity g changes by g? (Hint: The new period T+T is obtained by substituting g+g for g: T+T=2Lg+g. To obtain an approximate expression, expand the factor (g+g)1/2 using the binomial theorem and keep only the first two terms: (g+g)1/2=g1/212g3/2g+ The other terms contain higher powers of g and are very small if g is small.) Express your result as the fractional change in period TT in terms of the fractional change gg.
Part B
A pendulum clock keeps correct time at a point where g=9.8000m/s2, but is found to lose 5.0 s each day at a higher elevation. Use the result of part (a) to find the approximate value of g at this new location.
Express your answer using five significant figures.
Part B
A pendulum clock keeps correct time at a point where g=9.8000m/s2, but is found to lose 5.0 s each day at a higher elevation. Use the result of part (a) to find the approximate value of g at this new location.
Express your answer using five significant figures.
Explanation / Answer
PART A)
Time period of the Pendulum is given by,
T =2pi *sqrt(l/g).
taking ln on both sides:
ln t =ln 2pi +(1/2)[lnl] - lng].
take diffrencials of both sides
dt/t =(1/2)[dl/l-dg/g.]
if length remains constant,
We can write it as -
dt/t = - dg/2g
T/T = - 1/2 * (g/g)
PART B)
T = 5.0s
T = 24*3600 s
g = 9.8 m/s^2
g = ?
g = - 2*g * T/T
g = - 2*9.80 * 5/(24*3600)
g = - 0.00113
g = 9.8 - 0.00113
g = 9.7988 m/s^2