A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is

Question

A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is (A) T=2π √(L/g) (B) T=2π √(L/√g2+(V4)R2) (C) T=2π √(L/√g2+(V2)R) (D) T=2π √(L/√g2-(V4)R2)

Tardigrade Admin · Accepted Answer

The bob of the pendulum has two accelerations:
(i) Centripetal acceleration,

a_{c} = \frac{V ^{2}}{R}

It acts horizontally
(ii) Acceleration due to gravity = g. It acts vertically downwards
The effective acceleration due to gravity.

g^{'} = g^{2} + a_{c}^{2}

= g^{2} + (\frac{v ^{2}}{R})^{2}

g^{2} + \frac{v ^{4}}{R ^{2}}

∴

Time period,

T = 2 π \frac{L}{g ^{'}} = 2 π

\frac{L}{g ^{2} + \frac{v ^{4}}{R ^{2}}}

Q. A simple pendulum of length $L$ and having a bob of mass $m$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$ . If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is

$T = 2 π \frac{L}{g}$

$T = 2 π \frac{L}{g ^{2} + \frac{V ^{4}}{R ^{2}}}$

$T = 2 π \frac{L}{g ^{2} + \frac{V ^{2}}{R}}$

$T = 2 π \frac{L}{g ^{2} - \frac{V ^{4}}{R ^{2}}}$

Q. A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is

T=2πgL​​

T=2πg2+R2V4​​L​​

T=2πg2+RV2​​L​​

T=2πg2−R2V4​​L​​

Q. A simple pendulum of length $L$ and having a bob of mass $m$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$ . If the pendulum makes small oscillations in a radial direction about its equilibrium position, its time period of oscillation is

$T = 2 π \frac{L}{g}$

$T = 2 π \frac{L}{g ^{2} + \frac{V ^{4}}{R ^{2}}}$

$T = 2 π \frac{L}{g ^{2} + \frac{V ^{2}}{R}}$

$T = 2 π \frac{L}{g ^{2} - \frac{V ^{4}}{R ^{2}}}$