The displacement of a particle varies according to the relation $x = 3 sin 100 t + 8 {cos}^{2} 50 t$ . Which of the following is/are correct about this motion .

the motion of the particle is not S.H.M.

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the amplitude of the S.H.M. of the particle is

5

units

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the amplitude of the resultant S.H. M. is

\sqrt{73}

units

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the maximum displacement of the particle from the origin is

9

units .

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Solution

The correct options are
B the amplitude of the S.H.M. of the particle is $5$ units
D the maximum displacement of the particle from the origin is $9$ units .
$x = 3 sin 100 t + 8 {cos}^{2} 50 t$
$x = 3 sin 100 t + 4 (1 - c o s 100 t)$
$x = 3 sin 100 t - 4 cos 100 t + 4$
$x = 3 sin 100 t - 4 cos 100 t + 4$
Maximum value of $(3 sin 100 t - 4 cos 100 t)$ is 5
Amplitude=5
the maximum displacement of the particle from the origin is $5 + 4$ = 9
Hence, Option B and D is answer.

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