   Chapter 13, Problem 37P

Chapter
Section
Textbook Problem

A clock is constructed so that it keeps perfect time when its simple pendulum has a period of 1.000 s at locations where g = 9.800 m/s2. The pendulum bob has length L = 0.248 2 m, and instead of keeping perfect time, the clock runs slow by 1.500 minutes per day. (a) What is the free-fall acceleration at the clock’s location? (b) What length of pendulum bob is required for the clock to keep perfect time?

(a)

To determine
The free fall acceleration at the location of the pendulum.

Explanation

Given info: For the clock to have perfect time, the period of the pendulum is 1.000s where free fall acceleration is g=9.800ms-2 . The length of the pendulum is 0.2482m . At the location of the pendulum the clock runs slow by 1.500min .

One day is equal to 86400s . So the pendulum will oscillate 86400s times per day if the clock shows perfect timing.

If the clock is slow by 1.500min per day which will be equal to,

1.500minday=1.500×60sday=90.00sday

So the pendulum will oscillate 8640090=86310 times per day.

The time period of a simple harmonic motion is defined as the time required for completing one complete oscillation.

T=timetakennumberofoscillation=86400s86310=1

(b)

To determine
The length of the bob required for the clock to keep perfect time.

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