A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is π m / s . Amplitude of the oscillation is [ sin 3 0 ∘ = cos 6 0 ∘ = 0.5

Question

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is π m / s . Amplitude of the oscillation is [ sin 3 0 ∘ = cos 6 0 ∘ = 0.5 , sin 6 0 ∘ = cos 3 0 ∘ = 3 ​ /2 ]

TestHub · Accepted Answer

The displacement of a particle performing S.H.M. starting from the equilibrium position is given by:

$x = A sin (ω t)$

The velocity of the particle is:

$v = \frac{d x}{d t} = A ω cos (ω t)$

Given that the time period $T = 12$ s, the angular frequency is:

$ω = \frac{2 π}{T} = \frac{2 π}{12} = \frac{π}{6}$ rad/s

At $t = 2$ s, the velocity is $v = π$ m/s. Substituting these values into the velocity equation:

$π = A (\frac{π}{6}) cos (\frac{π}{6} \times 2)$

$π = A (\frac{π}{6}) cos (\frac{π}{3})$

Since $cos (\frac{π}{3}) = cos 6 0^{\circ} = 0.5 = \frac{1}{2}$ :

$π = A (\frac{π}{6}) (\frac{1}{2})$

$π = A (\frac{π}{12})$

$A = 12$ m

Answer: 12 m

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