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The displacement-time relation for a par...

The displacement-time relation for a particle can be expressed as `x=0.5[cos^(2)(npit)-sin^(2)(npit)]`. This relation shows that

A

the particle execute SHM with amplitude 0.5 m

B

the particle execute SHM with a frequency n times that of a second pendulum

C

the particle executes SHM and the velocity in its mean position is `(3.142n)m*s^(-1)`

D

the particle does not execute SHM at all.

Text Solution

Verified by Experts

The correct Answer is:
A, C
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Knowledge Check

  • The displacement of a particle in SHM varies according to the relation x=4(cospit+sinpit) . The amplitude of the particle is

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    `-4`
    B
    4
    C
    `4sqrt2`
    D
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    D
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  • The displacement y of a particle executing periodic motion is given by y=4cos^2(t/2)sin (1000t) . This expression may be considered to be a result of the superposition of how many independent harmonic motion?

    A
    five
    B
    two
    C
    three
    D
    four

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