Class 11th

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.

(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence it is a periodic motion. However, this motion is not simple harmonic, as earth does not have to and fro motion about its axis.

(b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillations. Their vibration is the superposition of individual simple harmonic motion of a number of different molecules. Hence, it is not simple harmonic, it is periodic.

(a) The swimmer's motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

(c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again, after an equal interval of time. Such motion is periodic.

(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence this motion is not a periodic.