Class 11th

(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

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

a)

The given situation is shown in the figure. Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path. A particle is in linear simple harmonic motion between the end points. At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point. Its acceleration is positive as it is directed along AO. Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point. Its acceleration is negative in this case as it is directed along B. Force is also negative as it is directed leftward.

(c)

The particle is executing a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)

The particle is moving towards point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle's velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving towards point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration and force are all positive.

(f)

 The particle is moving towards point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle's velocity and acceleration, and the force on it are all negative.

(a) The given function is sin $\omega$ t – cos $\omega$ t

= $\surd 2\left[\frac{1}{\surd 2}\sin ?\omega t-\frac{1}{\surd 2}\cos ?\omega t\right]$ = $\surd 2\left[\frac{1}{\surd 2}\sin ?\omega t*\cos ?\frac{\pi }{4}-\frac{1}{\surd 2}\cos ?\omega t*\sin ?\frac{\pi }{4}\right]$

= $\surd 2sin\left[\omega t-\frac{\pi }{4}\right]$ )

This function represents SHM as it can be written in the form: a sin ( $\omega$ t + $\theta$ )

Its period is $\frac{2\pi }{\omega }$ . It is periodic, but not SHM

(b) Sin³ $\omega$ t = $\frac{1}{2}\left[3\sin ?\omega t-\sin ?3\omega t\right]$

The terms $\sin ?\omega$ t and $\sin ?3\omega$ t individually represents simple harmonic motion. However, the superposition of two SHM is periodic but not simple harmonic.

(c) The given function is 3 cos ( $\pi$ /4 – 2 $\omega$ t) = -3cos(2 $\omega$ t - $\pi$ /4)

This function represents simple harmonic motion because it can be written in the form: a cos( $\omega t+\theta$ ), its time period is $\frac{2\pi }{2\omega }$ = $\frac{\pi }{\omega }$ . Periodic but not SHM

(d) The given function is cos $\omega$ t + cos 3 $\omega$ t + cos 5 $\omega$ t – Each individual cosine function represents SHM, however the superposition of three SHM is periodic but not simple harmonic.

(e) The given function exp (– $\omega$ ²t²) is an exponential function. Exponential functions do not repeat themselves, hence non-periodic motion.

(f) The given function 1 + $\omega$ t + $\omega$ ²t² is non-periodic.