Question

At what rate will a pendulum clock run on Titan, where the acceleration due to gravity is 1.37 m/s2, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock's hour hand to make one revolution on Titan.

Answer 1

`T_t =2.67\ hours`

Step-by-step explanation:

given,

acceleration due to gravity in titan = 1.37 m/s²

period of oscillation on earth

`T_e = 2\pi\sqrt{(L)/(g_e)}`

L is length of the pendulum

Period of oscillation on Tital

`T_t = 2\pi\sqrt{(L)/(g_t)}`

dividing the above equation

`(T_t)/(T_e) = \sqrt{(g_e)/(g_m)}`

time taken by hour hand for one revolution is equal to 1 hr

`T_t= T_e\sqrt{(g_e)/(g_m)}`

`T_t= 1 \sqrt{(9.8)/(1.37)}`

`T_t =2.67\ hours`

Answer 2

The time period on Titan is 2.67 hours.

Step-by-step explanation:

Given that,

Acceleration = 1.37 m/s²

Time on earth = 1 hour

Period of oscillation on earth

`T_(e)=2\pi\sqrt{(L)/(g_(e))}`

Where, L = length of the pendulum

The length of pendulum will be the same on the Titan.

The period of oscillation on Titan

`T_(t)=2\pi\sqrt{(L)/(g_(t))}`

We need to calculate the time period on Titan

`(T_(t))/(T_(e))=\frac{2\pi\sqrt{(L)/(g_(t))}}{2\pi\sqrt{(L)/(g_(e))}}`

`(T_(t))/(T_(e))=\sqrt{(g_(e))/(g_(t))}`

`T_(t)=T_(e)\sqrt{(g_(e))/(g_(t))}`

Put the value into the formula

`T_(t)=1*\sqrt{(9.8)/(1.37)}`

`T_(t)=2.67\ hours`

Hence, The time period on Titan is 2.67 hours.