wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

A particle executes SHM, with an amplitude of 10 cm. When the particle is at 6 cm from the mean position, the magnitude of its velocity is equal to twice of its acceleration. Find the time period of SHM(in seconds)

A
4.1 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9.42 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1.17 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
14.8 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 9.42 s
Given,
Amplitude (A)=10 cm
Position of particle from mean position (x)=6 cm
Since, particle executes SHM, we can use
v=ωA2x2 ... (1)
a=ω2x ... (2)
from (1) and (2) in |v|=2|a| (Given in the question)
we get
ωA2x2=2ω2x
ω=A2x24x2
Time period (T)=2πω
=2π4x2A2x2=2×3.14×2×6(10)2(6)2
=2×3.14×2×68
=9.42 s

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon