physics ncert solutions class 11th

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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.

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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.

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(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

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(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.

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(a, b, d) a) total mechanical energy of the body at any time t is

E= $\frac{1}{2}m$ w²a²

KE=1/2mv²= $\frac{1}{2}m[{\frac{dx}{dt}]}^2$

K_max= ½ mw²a²=E

b) K= ½ mw²a²cos²wt

for a cycle value of coswt is =1/2

 = 1/4 mw²a²= kmax/2

c) v=dx/dt = a coswt

V_maen =V_max+V_min/2

= aw-aw/2=0

d)V_rms= $\sqrt[]{\frac{v_1^2+v_2^2}{2}}=\sqrt[]{\frac{0+a^2{w}^2}{2}}=\frac{aw}{\sqrt[]{2}}$

V_rms=V_max/    $\sqrt[]{2}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) At t= 3T/4, the displacement of the particle is zero. Hence particle executing SHM will be at mean position i.e x=0 acceleration is zero and force is also zero.

At t= 4T/3, displacement is maximum i.e extreme position, so acceleration is maximum

At t = T/4 corresponds to mean position, so velocity will be maximum at this position.

At t= T/2 corresponds to extreme position so KE =0 and PE =maximum.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d) x=asinwt

V=dx/dt=awcoswt

A=dv/dt=-aw²sinwt

Force = mass $*acceleration$ = -mw²x

So force is directly proportional to displacement.