Question

The equation of a particle executing SHM is given by $x=150\sin (\frac{\pi }{3}t+\frac{\pi }{2})$, where the amplitude is in $\text{metres}$ and angular frequency is in $\text{rad/s}$. Find the maximum velocity $\left(v_{max}\right)$ and maximum acceleration $\left(a_{max}\right)$ attained by the particle.

• A. $v_{max}=50\pi \text{m/s},a_{max}=\frac{50{\pi }^2}{3}{\text{m/s}}^2$
• B. $v_{max}=100\pi \text{m/s},a_{max}=\frac{50{\pi }^2}{3}{\text{m/s}}^2$
• C. $v_{max}=50\pi \text{m/s},a_{max}=\frac{100{\pi }^2}{3}{\text{m/s}}^2$
• D. None of the above

Answer 1

The correct option is A $v_{max}=50\pi \text{m/s},a_{max}=\frac{50{\pi }^2}{3}{\text{m/s}}^2$

Given,
$x=150\sin (\frac{\pi }{3}t+\frac{\pi }{2})$
Comparing it with $x=A\sin (\omega t+\varphi )$, we get
$A=150\text{m},\omega =\frac{\pi }{3}\text{rad/sec},\varphi =\frac{\pi }{2}\text{rad}$
$V_{max}=A\omega =150\times \frac{\pi }{3}=50\pi \text{m/s}$
$a_{max}={\omega }^{2}A={\left(\frac{\pi }{3}\right)}^2\times 150=\frac{{\pi }^2}{9}\times 150=\frac{50{\pi }^2}{3}{\text{m/s}}^2$
Thus, option (a) is the correct answer.