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

A 100 gm block is connected to a horizontal massless spring of force constant 25.6 N/m and is free to oscillate on a horizontal frictionless surface. The block is displaced on a horizontal frictionless surface. The block is displaced by 3 cm along positive x - direction and released with an initial velocity of vi=163 cm.

Express position of block as function of time.


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


So far tile is at x=3 cm

And has v=163 cm/s

Hmmmm....... So where will the particle be if its seen as a projection of a particle doing uniform

Circular motion.

Since the particle is coming back it will be at B

Keeping this in mind Let's start.

Given m=0.1 kg

K=25.6 N/m

We know k=mω2ω=16

Particle is at x=3

v=163

From the above diagram we know ψ is greater then 90


flag
Suggest Corrections
thumbs-up
2
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The General Expression
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon