Question

The kinetic energy of a particle vibrating in SHM is 4J when it passes the mean position. If the mass of the body is 2kg,and the amplitude is 1m,calculate its time period

3.142s
2.142s
3.242s
3.542s

Answer 1

the answer is... 142s .

Answer 2

Answer:

The correct answer is $3.142 s$.

Explanation:

Concept:

A very significant category of periodic oscillation is known as simple harmonic motion, where the acceleration is proportional to the displacement (x) from equilibrium, in the direction of the equilibrium position. SHM is a periodic, to and fro shifting rotation about the mean position of the body. The restoring force on an oscillating body is geared toward its mean direction since it is directly proportional to its displacement.

Given:

When a particle passes the mean position while vibrating in SHM, its kinetic energy $=4J$

Mass of the body $=2kg$

The amplitude $= 1m$

To find:

We have to calculate its time period.

Solution:

By applying formula we get,

$&E^{\circ} K=\frac{1}{2} m \omega^{2} A^{2}$

$&E^{\circ} K=4 J \\m=2 K g\\A=1 m$

$&E=E^{\circ} K\\=\frac{1}{2} m \omega^{2} A^{2}$

$&\omega=\sqrt{\frac{2 E}{m A^{2}}} \text \\\\{ or } \\T=2 \pi A \sqrt{\frac{m}{2 E}}$

$&T=2 \pi \times 1 \times \sqrt{\frac{2}{2 \times 4}}\\\pi=3.142 s$

#SPJ3