Correct Answer is Option (C)

Equation of SHM is given by $x=A\sin \left(\omega t+\delta \right)$
$\left(\omega t+\delta \right)$ is called phase.

When x = $\frac{A}{2}$, then

$\sin \left(\omega t+\delta \right)=\frac{1}{2}$

$\Rightarrow \omega t+\delta =\frac{\pi }{6}$

$\Rightarrow {\varphi }_1=\frac{\pi }{6}$

For second particle,


${\varphi }_2=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$

$\therefore$ $\varphi ={\varphi }_2-{\varphi }_1$

$=\frac{4\pi }{6}=\frac{2\pi }{3}$

Correct Answer is Option (C)

y = sin$\omega$t – cos$\omega$t

$=\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin \omega t-\frac{1}{\sqrt{2}}\cos \omega t\right]$

$=\sqrt{2}\sin \left(\omega t-\frac{\pi }{4}\right)$

It represents a SHM with time period, $T=\frac{2\pi }{\omega }$

$y={\sin }^3\omega t=\frac{1}{4}\left[3\sin \omega t-\sin 3\omega t\right]$

It represents a periodic motion with time period

$T=\frac{2\pi }{\omega }$ but now SHM.

$y=5\cos \left(\frac{3\pi }{4}-3\omega t\right)$

$=5\cos \left(3\omega t-\frac{3\pi }{4}\right)$   $[\because \cos \left(-\theta \right)=\cos \theta ]$

It represents a SHM with time period, $T=\frac{2\pi }{3\omega }$

$y=1+\omega t+{\omega }^2{t}^2$

It represents a non-periodic motion. Also it is not physically acceptable as y $\to$ $\mathrm{\infty }$ as t $\to$ $\mathrm{\infty }$.

Correct Answer is Option (C)

$x=a{\sin }^2\omega t=a\left(\frac{1-\cos 2\omega t}{2}\right)$

   $(\because \cos 2\theta =1-2{\sin }^2\theta )$

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

$\therefore$ Velocity, $v=\frac{dx}{dt}=\frac{2\omega a\sin 2\omega t}{2}=\omega \alpha \sin 2\omega t$

Acceleration, $a=\frac{dv}{dt}=2{\omega }^{2}a\cos 2\omega t$

For the given displacement x = $\alpha \sin 2\omega t$,

$a\propto -x$ is not satisfied.

Hence, the motion of the particle is non simple harmonic motion.

Note : The given motion is a periodic motion with a time period

$T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$

Correct Answer is Option (C)

Speed $v=\omega \sqrt{a^2-x^2},x=\frac{a}{2}$

$\therefore$ $v=\omega \sqrt{a^2-\frac{a^2}{4}}=\omega \sqrt{\frac{3a^2}{4}}$

$=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}$

Correct Answer is Option (C)

Simple harmonic motion is defined as follows

Acceleration $\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}x$

The negative sign is very important in simple harmonic motion. Acceleration is independent of any initial displacement of equilibrium position.

Then acceleration = $-{\omega }^{2}x$.

Correct Answer is Option (D)

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e. $a\propto {\omega }^2$

$\therefore$ $\frac{a_1}{a_2}=\frac{{\omega }_1^2}{{\omega }_2^2}=\frac{{\left(100\right)}^2}{{\left(1000\right)}^2}=\frac{1}{100}$

$\Rightarrow$ a₁ : a₂ = 1 : 10².

Correct Answer is Option (B)

Displacement from the mean position

$y=a\sin \left(\frac{2\pi }{T}\right)t$

According to problem y = a/2

$a/2=a\sin \left(\frac{2\pi }{T}\right)t$

$\Rightarrow \frac{\pi }{6}=\left(\frac{2\pi }{T}\right)t\Rightarrow t=T/12$

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

Correct Answer is Option (D)

Let y = Asin$\omega$t

$\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sin \left(\omega t+\frac{\pi }{2}\right)$

Acceleration = $-A{\omega }^2\sin \omega t$

The phase difference between acceleration and velocity is $\pi /2$.

Correct Answer is Option (A)

In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.

Correct Answer is Option (D)

In S.H.M, $v_{max}=A\omega =A\left(2\pi f\right)$

$f=\frac{v_{max}}{2\pi A}=\frac{31.4}{2\left(3.14\right)\times 5}=1$ Hz per sec.

Correct Answer is Option (D)

In simple harmonic motion velocity
$=A\omega \sin \left(\omega t+\pi /2\right)$



acceleration $=A{\omega }^2\sin \left(\omega t+\pi \right)$ from this we can easily find out that when v is maximum, then $a$ is zero.