wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

The equation of a particle executing SHM is x=(5m)sin[(π s1)t+π3]. What is its amplitude, time period, maximum speed and velocity. At t = 1s?


A

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B


A typical equation representing SHM looks like x=A sin(ω t+ψ0)

Where x represents displacement

ω represents angular frequency

ψ0 represents initial phase

A represents amplitude

In the question we have x=(5m)sin[(π s1)t+π3]

So let's compare both! In a simple glance one can see

A=5m, ω=π s1 and ψ0=π3

Time period 2πω=2ππ=2s

Now what about velocities?

Let's differentiate the equation dxdt=(5m)π cos{(π s1)t+π3}

v=(5π)mcos{(π s1)t+π3}

We know that dxdt means velocity. To find the maximum velocity we can argue that cos has maximum value of 1 so Vmax=5π ms1. To

find velocity at t = 1s, let's substitute for t in velocity equation we get V1s=5π ms12.


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The General Expression
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon