Correct Answer is Option (A)

Here,

$x=x_0\cos \left(\omega t-\pi /4\right)$

$\therefore$ Velocity, $v=\frac{dx}{dt}=-x_0\omega \sin \left(\omega t-\frac{\pi }{4}\right)$

Acceleration,

$a=\frac{dv}{dt}=-x_0{\omega }^2\cos \left(\omega t-\frac{\pi }{4}\right)$

$=x_0{\omega }^2\cos \left[\pi +\left(\omega t-\frac{\pi }{4}\right)\right]$

$=x_0{\omega }^2\cos \left(\omega t+\frac{3\pi }{4}\right)$ $...\left(1\right)$

Acceleration, $a=A\cos \left(\omega t+\delta \right)$ $...\left(2\right)$

Comparing the two equations, we get

$A=x_0{\omega }^2$ and $\delta =\frac{3\pi }{4}.$

Correct Answer is Option (B)

Here, $x=2\times {10}^{-2}\cos \pi t$

$\therefore$ $v=\frac{dx}{dt}=2\times {10}^{-2}\pi \sin \pi t$

For the first time, the speed to be maximum,

$\sin \pi t=1$ or, $\sin \pi t=\sin \frac{\pi }{2}$

$\Rightarrow \pi t=\frac{\pi }{2}$ or, $t=\frac{1}{2}=0.5\sec .$

Correct Answer is Option (A)

Maximum velocity,

$v_{max}=a\omega ,v_{max}=a\times \frac{2\pi }{T}$

$\Rightarrow T=\frac{2\pi a}{v_{max}}=\frac{2\times 3.14\times 7\times {10}^{-3}}{4.4}\approx 0.01s$

Correct Answer is Option (A)

y = sin²$\omega$t

= $\frac{1-\cos 2\omega t}{2}$

$=\frac{1}{2}-\frac{1}{2}\cos 2\omega t$

$\therefore$ Angular speed = 2$\omega$

$\therefore$ Period (T) = $\frac{2\pi }{angularspeed}$ = $\frac{2\pi }{2\omega }$ = $\frac{\pi }{\omega }$

So it is a periodic function.

As y = sin²$\omega$t

$\frac{dy}{dt}$ = 2$\omega$sin$\omega$t cos$\omega$t = $\omega$ sin2$\omega$t

$\frac{d^{2}y}{d{t}^2}$ = $2{\omega }^2$ cos2$\omega$t which is not proportional to -y.

Hence it is is not SHM.

Correct Answer is Option (B)

$v_1=\frac{d{y}_1}{dt}=0.1\times 100\pi \cos \left(100\pi t+\frac{\pi }{3}\right)$

$v_2=\frac{d{y}_2}{dt}=-0.1\pi sin\pi t=0.1\pi cos\left(\pi t+\frac{\pi }{2}\right)$

$\therefore$ Phase diff. $={\varphi }_1-{\varphi }_2=\frac{\pi }{3}-\frac{\pi }{2}=\frac{2\pi -3\pi }{6}=\frac{\pi }{6}$