Physics, asked by vaidiksharma0004, 1 year ago

A particle executes S.H.M with a time period of 2S and amplitude 5 cm. Find
al Displacement b) Velocity​

Answers

Answered by Shubhendu8898
1

Answer: y =5sin(πt)

              v = 5π cos(πt)  

Explanation:

Given that,

Time period(T) = 2 seconds

Amplitude(a) = 5 cm

Since the particle executes SHM,

Phase(Ф) = 0

We know that equation of displacement of particle in SHM is,

y=a\sin(\omega t+\phi)\\\;\\y=5\sin(\omega t+0)\\\;\\y=5\sin{\frac{2\pi}{T}t}\;\;\;\;\;\;\;\;\;\;(\because\omega=\frac{2\pi}{T})\\\;\\y=5\sin(\frac{2\pi}{2}t)\\\;\\y=5\sin\pi t

Differentiating this equation with respect to t,

\frac{dy}{dy}=5\pi\cos\pi t\\\;\\v=5\pi\cos\pi t\;\;\;\;\;\;(\because \frac{dy}{dt}=v)

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