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A particle moves with a simple harmonic ...

A particle moves with a simple harmonic motion. If the velocities at distance of 4cm and 5cm from the equilibrium position are 13 cm per second and 5 cm per second respectively, find the period and amplitude.
[Hint: Use the formula v=`omegasqrt(a^(2)-x^(2)).`]

Text Solution

Verified by Experts

The correct Answer is:
1.57s,5.15cm
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Knowledge Check

  • A particle executing harmonic motion is having velocities v_1 and v_2 at distances is x_1 and x_2 from the equilibrium position. The amplitude of the motion is

    A
    `sqrt((v_1^2+x_2-v_2^2x_1)/(v_1^2-v_2^2))`
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    `sqrt((v_1^2x_1^2-v_2^2x_2^2)/(v_1^2+v_2^2))`
    C
    `sqrt((v_1^2x_2^2-v_2^2x_1^2)/(v_1^2-v_2^2))`
    D
    `sqrt((v_1^2x_2^2+v_2^2x_1^2)/(v_1^2+v_2^2))`

  • A particle vibrating simple harmonically has an acceleration of 16 cm s^(-2) when it is at a distance of 4 cm from the mean position. Its time period is

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    B
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    C
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    D
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  • A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm . At what displacement from the equilibrium position, is its energy half potential and half kinetic

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    B
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    C
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    D
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