Subjective Type

A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At $$t=0$$ it is at position $$x = 5 cm$$ going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at $$t = 4 s$$.

Solution

Given $$r=10cm$$
At $$t=0,x=5cm$$
$$T=6sec$$

So, $$w=\dfrac{2\pi}{T}=\dfrac{2\pi}{6}=\dfrac{\pi}{3}\sec^{-1}$$
At, $$t=0,x=5cm$$

So, $$5=10\sin(w\times 0+\phi)=10\sin\phi$$ [$$y=r\sin\,wt$$]
$$\sin\phi=1/2\Rightarrow \phi =\dfrac{\pi}{6}$$

$$\therefore$$ Equation of displacement $$x=(10cm)\sin\left(\dfrac{\pi}{3}\right)$$
At $$t=4\,\,second$$

$$x=10\sin\left[\dfrac{\pi}{3}\times 4+\dfrac{\pi}{6}\right]=10\sin\left[\dfrac{8\pi+\pi}{6}\right]$$

$$=10\sin\left(\dfrac{3\pi}{2}\right)=10\sin\left(\pi+\dfrac{\pi}{2}\right)=-10\sin \left(\dfrac{\pi}{2}\right)=-10$$
Acceleration $$a=-w^2x=-\left(\dfrac{\pi^2}{9}\right)\times (-10)=10.9\approx 0.11cm/sec$$

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