Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10-2cos(πt)m . The time at which the maximum speed first occurs is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

0.5 seconds

b

1 second

c

1.5 seconds

d

2 seconds

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Given displacement x=2×10-2cos(πt)  differentiate above equation with respect to time velocity =v=dxdt=−2×10−2πsin(πt) For the first time when V=Vmax velocity is maximum when sinπt=1=sin π2 πt= π2 t=12=0.5second***Alternate solution: The given equation shows that the particle is at extreme position .  since πt=ω ⇒πt=2πT  time period is equal to 2 sec and the particle will take a time equal to   T4=0.5sec                                                                                                                            to reach the mean position where the velocity is maximum.
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring