Class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(b) y = sin³wt

= [3sinwt-4sin3wt]/4

dy/dt= [ $\frac{d}{dt}\left(3sinwt\right)-\frac{d}{dt}\left(4sin3wt\right)$ ]/4

4dy/dt=3wcoswt-4 [3wcoswt]

4 $*\frac{d^{2}y}{d{t}^2}=-3w^{2}sinwt+12wsin3wt$

$\frac{d^{2}y}{d{t}^2}$ = - $\frac{3w^{2}sinwt+12w^{2}sin3wt}{4}$

$\frac{d^{2}y}{d{t}^2}$  is not proportional to y. hence it is not SHM.

This is a multiple choice answer as classified in NCERT Exemplar

Velocity =dy/dt= $\frac{d}{dt}\left. [3cos\left(\frac{\pi }{4}-2wt\right)\right]$

= 3 (-2w) [-sin ( $\frac{\pi }{4}-2wt$ )]

= 6wsin $\left. [\frac{\pi }{4}-2wt\right]$

Acceleration a = dv/dt= $\frac{d}{dt}\left. [6wsin\left(\frac{\pi }{4}-2wt\right)\right]$

= -4w² [3cos ( $\frac{\pi }{4}-2wt$ )]

A = -4w²y hence acceleration is directly proportional to displacement so it follows SHM

w'= 2w

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

T'= 2 $\pi /2w$ = $\frac{\pi }{w}$

This is a short answer type question as classified in NCERT Exemplar

By considering the diagram

_{$\theta$ 1}= ${\theta }_{o}sin(wt+{\varnothing }_1)$

_{$\theta$ 2}= ${\theta }_o\mathrm{s}\mathrm{i}\mathrm{n}?\left(wt+{\varnothing }_2\right)$

As it is clear given that amplitude time period being equal but phases being different. Now for first pendulum at any time t

_{$\theta$ 1}= $\theta$ ₂

So sin $\frac{\pi }{2}$ = sin(wt+ ${\varnothing }_1$ )

wt+ ${\varnothing }_1=\frac{\pi }{2}$

where $\theta$ _o=2^o is the angular amplitude of first pendulum . for the second pendulum , the angular displacement is one degree , therefore $\theta$ ₂= $\frac{{\theta }_0}{2}$  and negative sign is taken to show for being left to mean position.

- $\frac{{\theta }_o}{2}={\theta }_{0}sin(wt+{\theta }_2)$

Sin(wt+ ${\varnothing }_2$ )=-1/2

So (wt+ $\theta$ ₂)=- $\frac{\pi }{6}or\frac{7\pi }{6}$

So by making their difference

(wt+ $\theta$ ₂)-( $wt+\theta$ ₁)=7 $\frac{\pi }{6}-\frac{\pi }{2}$ =4 $\frac{\pi }{6}$

( $\theta$ _{2- $\theta$ 1})= 120⁰

This is a short answer type question as classified in NCERT Exemplar

Consider the diagram of the spring block system. It is a SHM with amplitude of 5cm about the mean position

Spring constant k=50N/m

Mass =2kg

Angular frequency w= $\sqrt[]{\frac{k}{w}}=\sqrt[]{\frac{50}{2}}=5rad/s$

Y (t)= Asin (wt+ $\varnothing$ )

Y (0)=Asin (w $*0+\varnothing$ )

sin $\varnothing$ =1 $\varnothing =\frac{\pi }{2}$

y (t)=Asin (wt+ $\frac{\pi }{2})$ = Acoswt

A=5cm w=5rad/s

Y=5sin5t

This is a short answer type question as classified in NCERT Exemplar

given potential energy associated with the field

U (x)=U_o (1-cos $\alpha x$ )

F=-dU (x)/dx

F=-d (U_o-U_ocos $\alpha x$ )=-U_o $\alpha$ sin $\alpha x$

F=-U_{o ${\alpha }^{2}x$}

F $\propto -x$

Motion is SHM for small oscillatons

F=-mw²X

Mw²=U_{o ${\alpha }^2$}

w²= $\frac{U_o{\alpha }^2}{m}$  , w= $\sqrt[]{\frac{U_o{\alpha }^2}{m}}$

T= $\frac{2\pi }{w}$ = $2\pi \sqrt[]{\frac{m}{U_o{\alpha }^2}}$

This is a short answer type question as classified in NCERT Exemplar

Let us assume that the required displacement be x

Potential energy of the simple harmonic oscillator =1/2 kx²

k= force constant=mw²

PE= ½ mw²x²

Maximum energy of oscillator

TE= ½ mw²A²

PE=1/2 TE

½ mw²x²= $\frac{1}{2}\left[\frac{1}{2}m{w}^2{A}^2\right]$

So x= $\sqrt[]{\frac{A^2}{2}}$ = $\pm \frac{A}{\sqrt[]{2}}$

This is a short answer type question as classified in NCERT Exemplar

As we know displacement y=sinwt-coswt

= $\sqrt[]{2}\left(\frac{1}{\sqrt[]{2}}sinwt-\frac{1}{\sqrt[]{2}}coswt\right)$

= $\sqrt[]{2}(cos\frac{\pi }{4}sinwt-sin\frac{\pi }{4}coswt)$

= $\sqrt[]{2}\left. [sin\left(wt-\frac{\pi }{4}\right)\right]$

To comparing with standard equation

Y= asin (wt+ $\varnothing$ )

So T=2 $\pi$ /w