wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The displacement (in m) equation of a particle is x=3 sin 2t+4 cos2t. The amplitude (in m) and maximum velocity (in m/s) will be respectively.

A
5,10
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3,2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4,2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3,4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 5,10
Given,
x=3 sin 2t+4 cos 2t
Rewriting the above equation as,
x=3 sin 2t+4 sin (2t+π2) ....(i)
sin(π2+θ)=cosθ
Equation (i) represents that displacement (x) of particle is due to superposition of two independent SHM, with a phase difference of π2
x=a1sinωt+a2sin(ωt+ϕ)

From the phases diagram given below:

The resultant amplitude of the particle will be given by,
a=a21+a22+2a1a2cosϕ
Where amplitude of individual SHM is, a1=3 m,& a2=4 m and phase difference ϕ=π2
a=a21+a22=32+42
a=5 m
Maximum velocity of the particle is given by,
(vmax)=aω
vmax=5×2=10 m/s

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon