Class 11th

This is a multiple choice answer as classified in NCERT Exemplar

Potential energy of a simple harmonic oscillator is = ½ kx²=1/2mw²x²

K=mw²

When x=0 PE=0

When x= $?A$ , PE=maximum

=1/2 mw²A²

KE of a simple harmonic oscillator =1/2 mv²

= 1/2 m [w $\sqrt[]{A^2-x^2}$ ] ²

= ½ mw² (A²-x²)

This is also parabola if plot KE against displacement x

KE= 0 at x= $?A$

KE=1/2mw²A² at x=0

Now total energy of the simple harmonic oscillator =PE+KE

= ½ mw²x²+1/2mw² (A²-x²)

TE= ½ mw²A²

So the curve according to that is

This is a multiple choice answer as classified in NCERT Exemplar

As we know x= acoswt

V =dx/dt= a (-sinwt)w=-wasinwt

V=-wasinwt

= wacos ( $\frac{\pi }{2}+wt$ )

Phase of velocity = $\frac{\pi }{2}+wt$

So difference in phse of velocity to that of phase of displacement = $\frac{\pi }{2}+wt-wt$ = $\frac{\pi }{2}$

This is a multiple choice answer as classified in NCERT Exemplar

As the particle on reference circle moves in anticlockwise direction. The projection will move from P to O towards left.

Hence in the position shown the velocity is directed from P' to P'' i.e from right to left . hence sign is negative.

This is a multiple choice answer as classified in NCERT Exemplar

In the diagram

the motion of a particle  executing SHM between A and B

Total distance travelled while it goes from A to B and returns to A is=AO+OB+BO+OA

= A+A+A+A=4A

So ratio of distance and amplitude =4A/A=4

This is a multiple choice answer as classified in NCERT Exemplar

As we know equation of SHM is x= Asinwt

V= dx/dt=Awsinwt

V_max=Awcoswt_max

= Aw

A=dv/dt=-wAwsinwt

= -w²Asinwt

A_max=-w²A

From above equations

$\frac{V_{max}}{A_{max}}$ =wA/w²A=1/w

This is a multiple choice answer as classified in NCERT Exemplar

The bob is displaced through some angle

The restoring force $\tau =-mgsin\theta$ if $sin\theta$ is small then it is $\theta$ only.

$\tau \approx -mg\theta$

So torque is directly proportional to angle.

So it clear from the above equation that its period will be harmonic

This is a multiple choice answer as classified in NCERT Exemplar

Acceleration is directly proportional to displacement.

The direction of acceleration is always towards the mean position that is opposite to displacement.

This is a multiple choice answer as classified in NCERT Exemplar

Consider the diagram in which the block is displaced right through x

The right spring gets compressed by x developing a restoring force kx towards left on the block. The left spring is stretched by an amount x developing a restoring force kx towards left on the block

Hence total force =kx+kx= 2kx towards left.

This is a multiple choice answer as classified in NCERT Exemplar

(i) In SHM y-t graph, zero displacement values correspond to mean position where velocity of the oscillator is maximum.

Where the crest and troughs represents extreme positions where displacement is maximum and velocity of the oscillator is minimum and is zero. Hence A

(ii) And also speed is maximum at mean position represented by B

This is a multiple choice answer as classified in NCERT Exemplar

(a, c, d) a) when the particle is 3 cm away from A going towards B, velocity is towards AB.i.e positive. SHM towards mean position so positive.

b) When the particle is at C velocity is towards B hence positive.

c) When the particle is 4 cm away from B is going towards A velocity is negative and acceleration towards mean position. Hence negative.

d) Acceleration is always towards mean position O. when the particle is at B acceleration and force are towards BA that is negative.