Question

The displacement $x$ of a simple harmonic oscillator varies with time $t$ as
$x\left(t\right)=0.5\sin (2\pi t+\frac{\pi }{4})$
What is the magnitude of maximum acceleration ?

• A. $\frac{8}{2}{\pi }^2\frac{\text{m}}{{\text{s}}^2}$
• B. $5{\pi }^2\frac{\text{m}}{{\text{s}}^2}$
• C. $2{\pi }^2\frac{\text{m}}{{\text{s}}^2}$
• D. $4{\pi }^2\frac{\text{m}}{{\text{s}}^2}$

Answer 1

The correct option is C $2{\pi }^2\frac{\text{m}}{{\text{s}}^2}$

Given that

$x\left(t\right)=0.5\sin (2\pi t+\frac{\pi }{4})$

$a=\frac{d^{2}x\left(t\right)}{d{t}^2}$

$\therefore \frac{dx}{dt}=2\pi \times 0.5\cos (2\pi t+\frac{\pi }{4})$

$=\pi \cos (2\pi t+\frac{\pi }{4})$

Now, $\frac{d^{2}x}{d{t}^2}=-2{\pi }^2\sin (2\pi t+\frac{\pi }{4})$

Maximum value at acceleration:

$\left|a\right|=\left|\frac{d^{2}x}{d{t}^2}\right|=|-2{\pi }^2\sin (2\pi t+\frac{\pi }{4})|$

$\therefore$$\left|a_{max}\right|=2{\pi }^2$

$[\because \sin (2\pi t+\frac{\pi }{4})=1ata=a_{max}]$

$\therefore \left|a_{max}\right|=2{\pi }^2\frac{\text{m}}{{\text{s}}^2}$

Hence, $\left(\text{C}\right)$ is correct.

Alternate solution: $x=A\sin (\omega t+\varphi )$ $a_{max}=A{\omega }^2$