Question

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 5 9.800 m/s2. The pendulum bob has length L 5 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?

Answer 1

Answer

Given,

Period of the Pendulum, T= 1 s

acceleration due to gravity, g = 9.8 m/s²

Length of Bob = 0.2482 m

Loss of time in the clock = 1.5 minutes/ day

Loss of time in 1 oscillation

   = $\dfrac{1.5\times 60}{24\times 60\times 60}$

   = 0.001042

Hence, time period, T = 1 + 0.001042

                                  T = 1.001042 s

a) Using Time period of oscillation formula

$T = 2\pi \sqrt{\dfrac{L}{g}}$

$1.001402 = 2\pi \sqrt{\dfrac{0.2482}{g'}}$

$g' = 9.77\ m/s^2$

Free fall acceleration at the clock location = 9.77 m/s²

b) Length of pendulum bob to keep it perfect time

 $1 = 2\pi \sqrt{\dfrac{L}{g}}$

 $1 = 2\pi \sqrt{\dfrac{L'}{9.77}}$

 $L' = 0.2475\ m$

Hence, Length of pendulum bob is equal to 0.2475 m.