Question

A particle executes simple harmonic motion with an amplitude of 10cm and time period 6s. At t=0 , it is at a distance if it is at point \[x=5cm\] going towards positive \[x\] direction.Write the equation of displacement (x) at time t . Find the magnitude of the acceleration of the particle at t =4s.

Answer 1

HintThe equation of displacement (along x axis is given by: \[x=A\cos \left( wt+\theta \right)\]) By putting initial condition values, the value of $\theta $ and hence the required equation is achieved. Velocity can then be calculated by differentiating x with respect to t and hence, acceleration can be calculated by differentiating \[v\] with respect to\[t\] .

Complete step by step solution
Equation of displacement along x axis is given by:
          \[\begin{align}
  & x=A\cos \left( \cot +\theta \right) \\
 & T=6\text{ sec (given)} \\
 & \therefore x=10\text{ cos}\left( \dfrac{2\pi }{T}+t+\theta \right) \\
 & x=10\text{cos}\left( \dfrac{2\pi }{6}+t+\theta \right) \\
 & \text{ At }t=0,x=5cm(\text{given)} \\
\end{align}\]
          \[\begin{align}
  & \therefore \text{5=}10\text{cos}\left( \dfrac{\pi }{3}\times 0+\theta \right) \\
 & \cos \theta =\dfrac{1}{2} \\
 & \theta -\dfrac{\pi }{3} \\
 & \therefore \text{Equation of displacement along }x\ \text{axis} \\
 & x=10\text{cos}\left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right) \\
\end{align}\]
Now , we have to calculate velocity
It is given by:
$\begin{align}
  & v=\dfrac{dy}{dx}=-10\sin \left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right)\times \dfrac{\pi }{3} \\
 & \text{acceleration }a=\dfrac{dy}{dx}=-10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{2\pi }{6}t+\dfrac{\pi }{3} \right) \\
 & \text{acceleration at }t=4 \\
\end{align}$
$\begin{align}
  & \text{ }a=-10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{2\pi }{6}\times 4+\dfrac{\pi }{3} \right) \\
 & \text{ }=10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( 300{}^\circ \right) \\
 & \text{ }=10{{\left( \dfrac{\pi }{3} \right)}^{2}}\cos \left( \dfrac{1}{2} \right) \\
 & \text{ }a=-5\cdot 48cm/{{\sec }^{2}} \\
 & \therefore \text{magnitude of acceleration is given by} \\
 & \text{ }a=-5\cdot 48cm/{{\sec }^{2}} \\
 & \text{ } \\
\end{align}$

Note Note that when displacement along x axis is given, the formula to be used is\[x=A\cos \left( wt+\theta \right)\]
Where displacement along y axis is given, then formula to be used is \[y=A\sin \left( wt+\theta \right)\]