physics ncert solutions class 11th

(a) The time period of a simple pendulum, T = 2 $\pi \sqrt{\frac{m}{k}}$

For a simple pendulum, k is expressed in terms of mass, m as : k $\propto m$ or $\frac{m}{k}$ = constant

Hence, the time period of a simple pendulum is independent of the mass of the bob. In the case of a simple pendulum, the restoring force acting on bob is given as F = -mg $\sin ?\theta$ , where

F = restoring force

m = mass of the bob

g = acceleration due to gravity

$\theta =\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}$

(b) For small $\theta ,$ sin $\theta$ $~\theta$ . For larger $\theta ,$ sin $\theta$ is greater than $\theta$ . This decreases the effective value of g.

Hence the time period increase as : T = 2 $\pi \sqrt{\frac{l}{g}}$ , where l is the length of the simple pendulum.

(c) The time shown by the wristwatch of a man falling from the top of the tower is not affected by the fall. Wrist watch works on spring action, independent of acceleration due to gravity.

(d) During free fall under gravity, the acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

(a) For figure (a) : When a force F is applied to the free end of the spring, an extension l is produced. For the maximum extension, it can be written as:

F – kl, where k is the spring constant.

For maximum =extension of the spring, l = $\frac{F}{k}$

For figure (b): The displacement (x) produced in this case is x = $\frac{1}{2}$

Net force F = +2kx = 2k $\frac{1}{2}$ . So l = $\frac{F}{k}$

(b) For figure (a) : For mass (m) of the block, force is written as : F = ma = m $\frac{d^{2}x}{d{t}^2}$ ,

where x is the displacement of the block in time t, then

m $\frac{d^{2}x}{d{t}^2}=-kx$ , it is negative because the direction of the elastic force is opposite to the direction of displacement.

$\frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{m}\right)x$ = ${-\omega }^2$ xwhere, ${\omega }^2$ = $\frac{k}{m}$ , $\omega \mathrm{i}\mathrm{s}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{s}\mathrm{c}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}.$

Time period of the oscillation, T = $\frac{2\pi }{\omega }$ = $\frac{2\pi }{\surd \frac{k}{m}}$ = 2 $\pi \sqrt{\frac{m}{k}}$

For figure (b) :

F = m $\frac{d^{2}x}{d{t}^2}$ = -2kx. It is negative because the direction of elastic force is opposite to the direction of displacement.

$\frac{d^{2}x}{d{t}^2}=-\left[\frac{2kx}{m}\right]x$ = - ${\omega }^2$ x, where angular frequency $\omega$ = $\sqrt{\frac{2k}{m}}$

Time period, T = $\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}$

(a) X = -2 sin( 3t + $\frac{\pi }{3}$ ) = +2cos ( 3t + $\frac{\pi }{3}$ + $\frac{\pi }{2})$ = 2 cos (3t + $\frac{5\pi }{6}$ )

when we compare this equation with standard SHM equation

x = Acos ( $\omega$ t + $\phi$ ), then we get

Amplitude A = 2 cm. Phase angle $\phi =\frac{5\pi }{6}$ = 150 $°$ , angular velocity $\omega$ = 3 rad/s

(b) X= cos ( $\frac{\pi }{6}-$ t) = cos ( $t-\frac{\pi }{6}$ )

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 1 cm. Phase angle $\phi =-\frac{\pi }{6}$ = - 30 $°$ , angular velocity $\omega$ = 1 rad/s

(c) X = 3sin (2 $\pi$ t + $\frac{\pi }{4}$ ) = -3cos $\left[\left(2\pi \mathrm{t}\mathrm{}+\mathrm{}\frac{\pi }{4}\right)+\mathrm{}\frac{\mathrm{\pi }}{2}\right]=-3\mathrm{c}\mathrm{o}\mathrm{s}?(2\pi \mathrm{t}+\mathrm{}\frac{3\pi }{4})$

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 3 cm. Phase angle $\phi =\frac{3\pi }{4}$ = 135 $°$ , angular velocity $\omega$ = $2\pi$ rad/s

(d) X = 2cos $\pi$ t

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 2 cm. Phase angle $\phi =0°$ , angular velocity $\omega$ = $\pi$ rad/s

(a) Time period, T = 2 s, Amplitude A = 3 cm

At time, t = 0, the radius vector makes an angle $\frac{\pi }{2}$ with the positive x-axis, i.e. phase angle $\phi$ = + $\frac{\pi }{2}$

Therefore, the equation of simple harmonic motion for the x-projection of the radius vector, at time t is given by the displacement equation:

x = Acos $\left. [\frac{2\pi t}{T}+\phi \right]$ = 3cos $\left. [\frac{2\pi t}{2}+\frac{\pi }{2}\right]$ = -3sin ( $\frac{2\pi t}{2}$ ) = -3sin $\pi t$ cm

(b) Time period, T = 4 s, Amplitude A = 2 m

At time, t = 0, the radius vector makes an angle $\pi$ with the positive x-axis, i.e. phase angle $\phi$ = + $\pi$

Therefore, the equation of simple harmonic motion for the x-projection of the radius vector, at time t is given by the displacement equation:

x = Acos $\left. [\frac{2\pi t}{T}+\phi \right]$ = 2cos $\left. [\frac{2\pi t}{4}+\pi \right]$ = -2cos ( $\frac{\pi }{2}t)$ m

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N/m and mass, m = 3 kg

Angular frequency,  $\omega =\sqrt{\frac{k}{m}}$ = $\sqrt{\frac{1200}{3}}$ = 20 rad/s

(a) When mass is at the mean position, displacement x = Asin $\omega t=2sin20t$ cm

(b) At the maximum stretched position, the mass is towards extreme right. Hence the initial phase is $\frac{\pi }{2}$

x = A sin ( $\omega t+\frac{\pi }{2})$ = 2sin (20t + $\frac{\pi }{2})$ = 2cos20t

(c) At the maximum compressed position, the mass is towards the extreme left

Hence, the initial phase is $\frac{3\pi }{2}$

x = Asin ( $\omega t+\frac{3\pi }{2})$ = 2 cos20t

Hence, the functions have the same frequency and amplitude, but different initial phases.

Initially at t =0, Displacement, x = 1 cm

Initial velocity, v = $\omega$ cm/s, Angular frequency, $\omega$ = $\pi$ rad/s

It is given that: x(t) = A cos ( $\omega$ t + $\phi$ )

Then 1 = A cos ( $\omega *0+\phi )$

A cos $\phi$ = 1 ….(i)

Velocity, v = $\frac{dx}{dt}$

$\omega =-A\omega \mathrm{s}\mathrm{i}\mathrm{n}?(\omega t+\phi )$

1 = -Asin( $\omega *0+\phi )=-Asin\phi$

Asin $\phi =-1$ ………(ii)

Squaring and adding equations (i) and (ii), we get

$A^2({sin}^2\phi +{cos}^2\phi )$ = 1+1

$A^2=2$

A = $\surd 2$ cm

$\tan ?\phi$ = -1

Then $\phi$ = $\frac{3\pi }{4}$ , $\frac{7\pi }{4}$ ,,,

SHM is given as X = Bsin( $\omega$ t + $\alpha$ )

Putting the given values in this equation, we get

1 = Bsin( $\omega *0+\alpha$ )

Bsin $\alpha$ =1…….(iii)

Velocity V = $\omega$ Bcos( $\omega$ t + $\alpha$ )

Substituting the given values, we get $\pi$ = $\pi$ Bcos $\alpha$ or B $\cos ?\alpha$ = 1 ….(iv)

Squaring and adding equations (iii) and (iv), we get

$B^2({sin}^2\alpha +{cos}^2\alpha )$ = 1+1

B = $\sqrt{2}cm$

Dividing equation (iii) by equation (iv), we get

$\frac{B\sin ?\alpha }{B\cos ?\alpha }$ = 1 or $\tan ?\alpha$ = 1

$\alpha$ = $\frac{\pi }{4}$ , $\frac{5\pi }{4}$