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The displacement of a simple harmonic os...

The displacement of a simple harmonic oscillator is given by `y= 0.40 sin (440t +0.61)`. For this, what are the value of amplitude.

Text Solution

Verified by Experts

Comparing `y= 0.40 sin (440t + 0.61)" with " y= A sin (omega t+ phi)`
Amplitude `A= 0.40 m`
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Knowledge Check

  • The displacement of a particle executing simple harmonic motion is given by y= A_(0) +A sin omega t+ B cos omega t . Then the amplitude of its oscillation is given by:

    A
    `A + B`
    B
    `A_(0) + sqrt(A^(2) + B^(2))`
    C
    `sqrt(A^(2) + B^(2))`
    D
    `sqrt(A_(0)^(2) + (A +B)^(2))`

  • The equation of simple harmonic is as following y(t) = 10 sin (20t+ 45^(@)) . Find the amplitude of SHM.

    A
    `a= 30`
    B
    `a= 20`
    C
    `a= 10`
    D
    `a=5`

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