Question

5) A block-spring system oscillates in a simple harmonic motion on a frictionless horizontal table. Its displacement varies with time according to x (t) = 0.2 cos (2t - n / 4). The earliest time the particle reaches position x = 0.1 m is

Answer 1

Explanations & Calculations

• This form of the equation exhibits that the time is measured with respect to a point located between the center & the amplitude.
• As usual the distance $\small x$ is measured with respect to the center.
• Then substituting the given value of x in the above equation, the time taken to reach that point can be calculated.

$\qquad\qquad \begin{aligned} \small 0.1&=\small 0.2\cos(2t-\frac{\pi }{4})\\ \small \frac{1}{2}=\cos\frac{\pi}{3}&=\small \cos(2t-\frac{\pi}{4})\\ \small \frac{\pi}{3}&=\small 2t-\frac{\pi}{4}\\ \small 2t&=\small \frac{7\pi}{12}\\ \small t&=\small \frac{7\pi}{24}\\ &=\small \bold{1.833\,s} \end{aligned}$ $\qquad\qquad \begin{aligned} \text{Extra }&\text{ Information}\\ \small x=0&\to t=\frac{T}{4}\\ \small \frac{\pi}{2}&=\small \Bigg(2\Big(\frac{T}{4}\Big)-\frac{\pi}{4}\Bigg)\\ \small \frac{T}{4}&=\small \frac{3\pi}{8}\approx 1.178\,s\\ \small T&=\small 4.712\,s \end{aligned}$

• Extra information,
• On comparison with the general form of the equation, that is $x(t)=A\cos(\omega t\pm\theta)$ , A=0.2 & $\omega =2$. Then the periodic time can be found to be,

$\qquad\qquad \begin{aligned} \small \omega&=\small \frac{2\pi}{T}=2\\ \small T&=\small \pi \approx3.142\,s \end{aligned}$