Question

Q1.A particle executes S.H.M with a time period of 2S and amplitude 5 cm. Find
al Displacement b) Velocity​.

Answer 1

Step-by-step explanation:

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,

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

Differentiating this equation with respect to t,

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