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PhysicsSHMMiscellaneousMedium2 minPYQ_2017

PhysicsMediumsingle choice

A particle performing S.H.M. starts from the equilibrium position and its time period is $16$ seconds. After $2$ seconds its velocity is $π m s^{- 1} .$ Amplitude of oscillation is $(\cos 45^{o} = \frac{1}{\sqrt 2})$

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Answer:

Solution:

Displacement of the particle $= x = A \sin ω t$
The velocity of the particle $= v = \frac{d x}{d t} = A ω \cos ω t$
$v = π m s^{- 1}, T = 16 s, ω = \frac{2 π}{T} = \frac{2 π}{16} = \frac{π}{8} rad s^{- 1}$
$∴ π = A \times \frac{π}{8} \times \cos \frac{π}{8} \times 2$
$∴ 1 = \frac{A}{8} \cos \frac{π}{4} = \frac{A}{8} . \frac{1}{\sqrt 2}$
$∴ A = 8 \sqrt{2} m$

Stream:MHT_CETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2017

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