The displacement of a particle executing SHM is given by y=5≤ft(4t+π/3) If T is the time period and the mass of the particle is 2g, the kinetic energy of the particle when t = T/4 is given by

Correct Answer is : 0.3 J0.3\;{J}0.3J

Correct Answer is : 0.3 J0.3\;{J}0.3J

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VITEEE 2009 Paper

Question : 8 of 120

Marks: +1, -0

y=5\sin\left(4t+\frac{\pi}{3}\right)

Solution:

Particle executing SHM.

Displacement

y=5\sin\left(4t+\frac{\pi}{3}\right)

...(i)

Velocity of particle

\frac{dy}{dt}=5\frac{d}{dt}\sin\left(4t+\frac{\pi}{3}\right)

=5\cos\left(4t+\frac{\pi}{3}\right)\cdot4

=20\cos\left(4t+\frac{\pi}{3}\right)

Velocity at

t=\frac{T}{4}

\left(\frac{dy}{dt}\right)_{t=\frac{T}{4}}=20\cos\left(4\times\frac{T}{4}+\frac{\pi}{3}\right)

\;u=20\cos\left(T+\frac{\pi}{3}\right)

..(ii)

Comparing the given equation with standard equation of SHM.

y=a\sin(\omega t+\phi)

We get, ω = 4

\;\omega=\frac{2\pi}{T}

⇒

\;T=\frac{2\pi}{\omega}

\;\;T=\frac{2\pi}{4}=\frac{\pi}{2}

Now putting value of T in Eq.(ii), we get

u=20\cos\left(\frac{\pi}{2}+\frac{\pi}{3}\right)=-20\sin\frac{\pi}{3}=-10\times\sqrt{3}

The kinetic energy of particle,

KE=\frac{1}{2}\mu^2=\frac{1}{2}\times2\times10^{-3}\times(-10\sqrt{3})^{2}

\;10^{-3}\times100\times3=0.3\;\mathrm{J}

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