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The resultant amplitude due to super pos...

The resultant amplitude due to super position of `x_(1)=sin omegat, x_(2)=5 sin (omega t +37^(@))` and `x_(3)=-15 cos omega t ` is `:`

A

17

B

21

C

13

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
`x_(1)=sin omegat,x_(2)=5 sin ( omegat+37^(@))`
`x_(3)=15 sin ( omegat-pi//2)`
By the phasor diagram `,`

Get the resultant of these 3 vectors as 13.
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x_(1) = 3 sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t Find amplitude of resultant SHM.

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Knowledge Check

  • Find the resultant amplitude of the following simple harmonic equations: x_(1)=5 sin omegat x_(2)=5 sin (omgat+53^(@)) x_(3)=-10 cos omegat

    A
    5
    B
    10
    C
    15
    D
    20

  • The resultant amplitude due to superposition of three simple harmonic motions x_(1) = 3sin omega t , x_(2) = 5sin (omega t + 37^(@)) and x_(3) = - 15cos omega t is

    A
    `18`
    B
    `10`
    C
    `12`
    D
    None of these

  • Find the resultant amplitude of the following simple harmonic equations : x_(1) = 5sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t

    A
    5
    B
    10
    C
    15
    D
    20

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