Home
Class 11
PHYSICS
Assertion : Aparticle is under SHM along...

Assertion : Aparticle is under SHM along the x - axis. Its mean position is `x = 2`, amplitude is `A = 2` and angular frequency `omega`. At `t = 0`, particle is at origin, then `x` - co-ordinate versus time equation of the particle will be `x = -2 cos omega t + 2`.
Reason : At `t = 0` , particle is at rest.

A

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

B

If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C

If Assertion is true, but the Reason is false.

D

If Assertion is false but the Reason is true.

Text Solution

Verified by Experts

The correct Answer is:
B
`y = (- A cos omega t)` at the particle starts from `- A`
So, displacement from the mean position will be `-A cos omega t`.

`:. x = y + 2`
`= -2 cos omega t + 2`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    DC PANDEY|Exercise Level 1 Single Correct|24 Videos

  • SIMPLE HARMONIC MOTION

    DC PANDEY|Exercise Level 1 Subjective|39 Videos

  • SIMPLE HARMONIC MOTION

    DC PANDEY|Exercise Miscellaneous Examples|8 Videos

  • SEMICONDUCTORS AND ELECTRONIC DEVICES

    DC PANDEY|Exercise More than One Option is Correct|3 Videos

  • SOLVD PAPERS 2017 NEET, AIIMS & JIPMER

    DC PANDEY|Exercise Solved paper 2018(JIPMER)|38 Videos

Similar Questions

Explore conceptually related problems

Equation motion of a particle is given as x = A cos(omega t)^(2) . Motion of the particle is

The motion of a particle is given x = A sin omega t + B cos omega t The motion of the particle is

A particle starts from rest at the extreme position and makes SHM with an amplitude A and angular frequency omega . Its displacement at time 't' is

Position of a particle varies as y = cos^(2) omega t - sin^(2) omega t . It is

A particle moves along the x-axis according to the equation x=a sin omega t+b cos omega t . The motion is simple harmonic with

A particle executes SHM along x-axis with an amplitude A, time period T with origin as the mean position. At t = 0, if the particle starts in the +ve x-direction from the origin the minimum time in which it will be at x=-A//2 will be

Two particles are in SHM with same amplitude A and same regualr frequency omega . At time t=0, one is at x = +A/2 and the other is at x=-A/2 . Both are moving in the same direction.

the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is

A x and y co-ordinates of a particle are x=A sin (omega t) and y = A sin(omegat + pi//2) . Then, the motion of the particle is

A particle moves on the x -axis according to the equation x=x_(0)sin2 omega t .The motion is simple harmonic ?