Question

A grandfather clock is based on the period of oscillation of a pendulum. The clock keeps perfect time on the first floor of the Sears tower but when carefully moved to the top floor h=436 m above the surface of the earth, the clock is found to be slow.

(a) What is the reason for this?

(b) How many seconds does the clock loose in one day on the top floor? (The radius of the earth is R=6.37. 106 m)

Answer #1

(a) time period of a simple pendulum= T = 2 pi L/g

L= length of pendulum and g= gravitational accelertion

as we move above the surface of earth, **value of g
decreses**, so the **time period of pendulum clock
increases**. hence the **clock slows**
down.

(b) T = 2 pi L/g ---------------------------- (1)

where g= 9.81 m/s^2 (on the surface) and T= 24 hours=86400 seconds

lets first find value of g at a height of 436 m above the earth,

new value of g is given by gn= GM/(R+h)^2

gn= (6.67x10^-11 * 5.98x10^24)/ (6.37x10^6 + 436)

gn= 9.78 m/s^2

New period= Tn= 2 pi L/gn --------------------------------(2)

we divide the equation 2 by equation 1

Tn/T= g/gn

Tn= T 9.81/9.78

Tn= 86400 *1.0015326

Tn= 86532 seconds

so seconds lost in one day= Tn- T

seconds lost= 86532-86400

**seconds lost= 132 s**

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