# Question d1df7

Dec 7, 2017

(a) 0.25 m

#### Explanation:

Time period = 1.6 s
Angular frequency, omega = (2π)/T = 3.927# rad s⁻¹

Since we know the object passes the equilibrium with a velocity of 1.0 m s⁻¹ that is the object's maximum velocity.

Use this equation to solve the problem:
${v}_{\max} = \omega A$

$A = {v}_{\max} / \omega = \frac{1.0}{3.927} = 0.2546$ m

So the answer is (a)

Dec 7, 2017

Option ( B)

#### Explanation:

Let the object $\textcolor{g r e e n}{P}$ executes SHM on Y-axis maintaining its equilibrium position at origin O . Given that the time period of oscillation is $T = 1.6$s. The angular velocity $\omega$ of the reference point $\textcolor{red}{R}$ associated with SHM undergoing uniform circular motion will be given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{1.6} \text{rad/s}$.

Considering that time count is started when the object is at origin. So we can write the equation for displacement $x$ in t sec as

$\textcolor{red}{y = a \sin \omega t \ldots . . \left[1\right]}$ , where $a$ is the amplitude of vibration

Here origin is also its equilibrium position and It returns repeatedly to this position after each complete oscillation for 1.6s.

So the velocity $v$ of the object after $t$ sec will be obtained by differentiating the above equation w r to t

$\textcolor{b l u e}{v = \frac{\mathrm{dy}}{\mathrm{dt}} = a \omega \cos \omega t \ldots \ldots . \left[2\right]}$

Now it is given that the velocity of the object after passing the equilibrium position by $0.2$s is $1 m \text{/s}$

So we can insert $t = 0.2$s, $v = 1 m \text{/s}$ in [2] to find out $a$

$v = a \omega \cos \omega t$

$1 = a \times \frac{2 \pi}{1.6} \times \cos \left(\frac{2 \pi}{1.6} \times 0.2\right)$

$\implies 1 = a \times \frac{2 \pi}{1.6} \times \cos \left(\frac{\pi}{4}\right)$

$\implies 1 = a \times \frac{2 \pi}{1.6} \times \frac{1}{\sqrt{2}}$

$\implies a = \frac{0.8 \times \sqrt{2}}{\pi} \approx 0.36 m$