Question

in 20sec two simple pendulum P and Q complete 9 and 7 oscillation resp on earth . On the moon where the acceleration due to gravity 1/6 th that on the earth the periods in ratio​

Answer 1

Answer:

A simple pendulum makes 20 oscillations in one second on the surface of the earth. Determine the time period of the simple pendulum on the surface of a planet where the acceleration due to gravity is one fourth of the acceleration due to gravity on the surface of the earth .

Answer 2

Given two pendulums oscillating with different time periods, Find the ratios of their time periods to that on moon  

Explanation:

• The time period of the oscillation of a simple pendulum of length 'L' and acceleration due to gravity acting on it 'g' is given by,     $T=2\pi\sqrt{\frac{L}{g} }$    
• Given the relation between acceleration due to gravity on earth and moon as,  $g_m=\frac{g_e}{6}$
• Let the time period on earth be denoted by 'Te' and on moon be denoted by ' Tm ' then from above points we have,     $\frac{T_m} {T_e}=\sqrt{\frac{g_e}{g_m} } \ \ \ \ \ \ \ ->T_m= T_e\sqrt{6}$      ---(a)
• The time period of both the pendulums on earth is, $T_{e1}=\frac{20}{9}\ s \ \ \ \ \ \ T_{e2}= \frac{20}{7} \ s$    
• Hence from (a) we get the time periods on the moon as,  $T_{m1}=T_{e1}\sqrt{6}\ \ \ \ \ \ \ \ \ \ ->T_{m1}= \frac{20\sqrt{6} }{9} \\T_{m2}=T_{e2}\sqrt{6} \ \ \ \ \ \ \ \ \ ->T_{m2}=\frac{20\sqrt{6} }{7}$  
• Now the ratio of the time periods of both the pendulums on earth and moon is, $T_{e1}:T_{e2}=T_{m1}:T_{m2}=7:9$