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The tota energy of a particle executing ...

The tota energy of a particle executing simple harmonic motion is `16J`. What will be total energy of particle if amplitude is halved and frequency is douboled?

A

`1J`

B

`64 J`

C

`16J`

D

`256J`

Text Solution

Verified by Experts

The correct Answer is:
C
`T.E. = (1)/(2) m omega^(2) A^(2)`
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Knowledge Check

  • The velocity of a particle executing simple harmonic motion is

    A
    `omega^(2)sqrt(A^(2)+x^(2))`
    B
    `omegasqrt(A^(2)-x^(2))`
    C
    `omegasqrt(A^(2)+x^(2))`
    D
    `omega^(2)sqrt(A^(2)-x^(2))`

  • The total energy of a particle executing simple garmonic motion is (x- displacement)

    A
    `prop x`
    B
    `prop x^(2)`
    C
    Independent of `x`
    D
    `prop x^(1//2)`

  • The phase of a particle executing simple harmonic motion is pi/2 when it has

    A
    Maximum velocity
    B
    Maximum acceleration
    C
    Maximum energy
    D
    Maximum displacement

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