Question

The bob of a simple pendulum performs S.H.M. with period 'T' in air and with period '$T_1$' in water. Relation between 'T' and '$T_1$' is (neglect friction due to water, density of the material of the bob is $\frac{9}{8}\times {10}^{3}kg/m^3,$ density of water = 1 g/cc)

• A. $T_1=3T$
• B. $T_1=2T$
• C. $T_1=T$
• D. $T_1=\frac{T}{2}$

Answer 1

The correct option is A $T_1=3T$

In air, time period of oscillation of pendulum will be $T=2\pi \left(\frac{l}{g}{)}^{(}\frac{1}{2}\right)$..........(1)

but in water $m{a}_{net}=F_{gravity}-F_{buoyancy}$

${\rho }_{bob}\times V_{bob}\times a_{net}={\rho }_{bob}\times V_{bob}\times g-{\rho }_{water}\times V_{bob}\times g$

This will give $a_{net}=\frac{({\rho }_{bob}-{\rho }_{water})}{{\rho }_{bob}}\times g$

$=(1-\frac{{\rho }_{water}}{{\rho }_{bob}})\times g$

$T_1=2\pi \left(\frac{l}{a_{n}et}{)}^{(}\frac{1}{2}\right)$.................(2)

On dividing (1) and (2)

we will get $T_1=3T$