First slide

Simple harmonic motion

Question

The total mechanical energy of a harmonic oscillator of amplitude 1 m and force constant 200 N/m is 100 J then

Moderate

A

The minimum PE is zero
B

the maximum PE is 100 J
C

the minimum PE is 50 J
D

the maximum PE is 50 J

Solution

given force constant k=200N/m
total energy TE=100 J
maximum kinetic energy is KE $= \frac{1}{2} {mw}^{2} (A^{2})$
maximum kinetic energy KE= $\frac{1}{2} k (A^{2})$
maximum kinetic energy KE $= \frac{1}{2} \times 200 \times 1^{2}$
$∴$ $K E_{m a x} = 100 J$
Total energy TE= $K E_{m a x} + P E_{m i n}$
100=100+ $P E_{m i n}$
$P E_{m i n}$ =0 J

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App