Test Index

JEE Advanced 2016 Paper 2

Section: Physics

Question : 14 of 54

Marks: +1, -0

x_0

x_0

x = x_0 + A

The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$ , whereas in thesecond case it remains unchanged
The final time period of oscillation in both the cases is same
The total energy decreases in both the cases
The instantaneous speed at $x_0$ of the combined masses decreases in both the cases.

Solution:

T_i 2\pi \sqrt{\frac{M}{K}}, T_f = 2\pi \sqrt{\frac{M+m}{K}}

Case (i):

M(A\omega) = (M+m)V

∴ Velocity decreases at equilibrium position

By energy conservation

A_f = A_i \sqrt{\frac{M}{m+m}}

Case (ii):No energy loss, amplitude remains same

At equilibrium

(x_0)

velocity

= A\omega

In both cases

\omega

decreases so velocity decreases in both cases

Go to Question:

Prev Question

Next Question