Question

a particle of mass 0.5 kg executes SHM .its period of oscillation is 3.14 sec a d its total energy is 0.04 j , its amplitude is​

Answer 1

Given:-

• a particle of mass 0.5 kg executes SHM.
• its period of oscillation is 3.14 sec.
• total energy is 0.04 j .

To Find:-

• Amplitude.

Solution:-

$\sf \: Total \: e nergy= K+U$

$K= \frac{1}{2} m {\omega}^{2} ( {a}^{2} - {y}^{2} )$

and

$U = \frac{1}{2} m {\omega}^{2} {y}^{2}$

$\therefore \: Total \: energy = \frac{1}{2} m {\omega}^{2} {a}^{2}$

$= > 0.04 = \frac{1}{2} \times0.5 \times {( \frac{2\pi}{t}) }^{2} \times {a}^{2}$

$= > 0.04 = \frac{1}{2} \times0.5 \times {( \frac{2 \times 3.14}{3.14}) }^{2} \times {a}^{2}$

$= > 0.04 = \frac{1}{2} \times 0.5 \times 4 \times {a}^{2}$

$= > 0.04 = {a}^{2}$

$= > \: a = 20cm$

∴The Amplitude is 20cm.