Question

The displacement of an oscillating particle varies with time (in seconds) according to the equations $y\left(cm\right)=sin\frac{\pi }{2}(\frac{t}{2}+\frac{1}{3})$. The maximum acceleration of the particle is approximately.

• A. $5.21cm/s^2$
• B. $3.62cm/s^2$
• C. $1.81cm/s^2$
• D. $0.62cm/s^2$

Answer 1

The correct option is D $0.62cm/s^2$

From the given equation of the oscillation, we obtain,

The angular frequency of the oscillation $\omega =\frac{\pi }{4}$

The amplitude of the particle is $a=1$

So, the maximum acceleration of the particle under oscillation is given as:

$a_{max}={\omega }^{2}a={\left(\frac{\pi }{4}\right)}^2\times 1$

$=0.62cm/se{c}^2$ [$\because a=1$]