Question

Problem 11.47

1. If a pendulum has a length L and you want to double its frequency, what should be its length in terms of L?

2. Suppose a pendulum has a length L and period T on earth. If you take it to a planet where the acceleration of freely falling objects is twenty times what it is on earth, what should you do to the length to keep the period the same as on earth?

3. If you do not change the pendulum's length in part (2), what is its period on that planet in terms of T?

Answer 1

1) The frequency of the pendulum is given by

w = sqrt(g/L)

new frequency, w' = 2w, we have to find new length, L'

for this new frequency,

w' = sqrt(g/L')

dividing both the equations

w/w' = sqrt(L'/L)

w/(2w) = sqrt(L'/L)

1/2 = sqrt(L'/L)

1/4 = L'/L

L' = L/4

2) Period of pendulum is,

T = 2*sqrt(L/g)

For new case,

T' = T;   g' = 20g; we have to find the new length; and, T' = 2*sqrt(L'/g')

equating both the equations

2*sqrt(L/g) = 2*sqrt(L'/g')

sqrt(L/g) = sqrt(L'/g')

L/g = L'/g'

Lg' = L'g

L*20*g = L'*g

L' = 20L

3) T' = 2*sqrt(L'/g')

Now, g' = 20g; L' = L

T' = 2*sqrt(L/20g)

T' = sqrt(1/20)* 2*sqrt(L/g)

T' = sqrt(1/20)* T