1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The total energy of the particle executing simple harmonic motion of amplitude A is 100 J. At a distance of 0.707 A from the mean position, its kinetic energy is

A
25 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
50 J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
100 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12.5 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
70 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 50 JWe know that Total energy E=12mω2A2=100JAnd, kinetic energy KE==12mω2(A2−y2)Now, position of particle from its mean position y=0.707A=A√2KE=12mω2⎡⎣A2−(A√2)2⎤⎦KE=12mω2[2A2−A22]KE=12mω2A22KE=1002J=50J

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Energy in SHM
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program