Question

A simple pendulum of length I 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, what will be its time period ?​

Answer 1

Answer :

• Length = l
• Mass = M
• Radius = R
• Speed = v
• Time Period = ?

━━━━━━━━━━━━━━━━━━

$\sf :\implies a = \sqrt{g^2+ {a_c}^2}$

$\qquad\quad\dag\small\sf {a_c}^2 = \dfrac{v^2}{R}$

$\sf :\implies a = \sqrt{g^2+ {\bigg\lgroup\dfrac{v^2}{R}\bigg\rgroup}^2}$

$\sf :\implies a = \sqrt{g^2+ {\bigg\lgroup\dfrac{v^4}{R^2}\bigg\rgroup}^2}$

• When T is the time period of thr oscillation then we know that,

$\sf :\implies T = 2\pi \sqrt{\dfrac{l}{a}}$

$\sf :\implies T = 2\pi \sqrt{\dfrac{l}{\sqrt{g^2+\dfrac{v^4}{R^2}}}}$

$\sf :\implies\pink{ T = 2\pi \sqrt{\dfrac{l}{g^2+\bigg\lgroup\dfrac{v^4}{R^2}\bigg\rgroup^{\frac{1}{2}}}}}$

Answer 2

Given:-

✦A simple pendulum of length I 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.

✦The pendulum makes small oscillations in a radial direction about its equilibrium position,

━━━━━━━━━━━━━━━━

To Prove:-

✦What will be its time period?..

━━━━━━━━━━━━━━━━

Solution:-

✦We know,

ar = v² / R

✦( ar - radial accleration )

✦( v - speed of the car on circular track )

✦( R - radius of the circular track )

➛Effective accleaccleration vector = - v² / R i - gj

➛Magnitude of effective acceleration = √ ( g² + v⁴ / R² )

We know,

➛=T = 2πr √(L / aeff)

➛=2πr √(L / √( g² + v⁴ / R² )

➛= 2πr √(L /( g² + v⁴ / R²)

➛The required time period is 2πr √(L /( g² + v⁴ / R²)

━━━━━━━━━━━━━━━━