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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

Figure 14.2. shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

(a) x (t) = B sin(2 $\frac{π t}{30}$ )

(b) x (t) = B cos(πt/15)

(c) x (t) = B sin( $\frac{π t}{15} + \frac{π}{2}$ )

(d) x (t) = Bcos ( $\frac{π t}{15} + \frac{π}{2}$ )

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Let angular velocity of the particle executing circular motion is w and when it is at O makes and angle $θ$

As $θ$ =wt

OR=OQCos (90- $θ$ )

= OQsin $θ$ =OQsinwt

=rsinwt

x=rsinwt=Bsinwt

= Bsin $\frac{2 π}{T} t$ =Bsin ( $\frac{2 π}{30} t$ )

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

Four pendulums A, B, C and D are suspended from the same

elastic support as shown in Fig. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,

(a) D will vibrate with maximum amplitude.

(b) C will vibrate with maximum amplitude.

(c) B will vibrate with maximum amplitude.

(d) All the four will oscillate with equal amplitude.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) A is given a transverse displacement. Through the elastic support the disturbance is transferred to all the pendulums.

A and C are having same length hence they will be in resonance, because their time period of oscillation.

2 π \sqrt[]{\frac{l}{g}}

hence frequency is same. So amplitude of A and C will be maximum.

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt.

(a) The motion is oscillatory but not S.H.M.

(b) The motion is S.H.M. with amplitude a + b.

(c) The motion is S.H.M. with amplitude a² + b².

(d) The motion is S.H.M. with amplitude $\sqrt[]{a 2 + b 2}$ .

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) y= asinwt + bcoswt

a=Asin $\emptyset$ and b= Acos $\emptyset$

a²+b²=A²sin^{2
$\emptyset$} +A²cos^{2
$\emptyset$}

A= $\sqrt[]{a^{2} + b^{2}}$

y=Asin $\emptyset s i n w t$ +Acos $\emptyset c o s w t$

= Asin (wt+ $\emptyset$ )

dy/dt= Awcos (wt+ $\emptyset$ )

$\frac{d^{2} y}{d t^{2}} = - A w^{2} s i n (w t + \emptyset) = - A y w^{2}$

So it is proportional to displacement . so follows SHM

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

A particle is acted simultaneously by mutually perpendicular simple hormonic motions x = a cos ωt and y = a sin ωt . The trajectory of motion of the particle will be

(a) An ellipse.

(b) A parabola.

(c) A circle.

(d) A straight line.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Y= asinwt

Squaring and adding above eqns

x²+y²=a², this is the equation of circle

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

Motion of an oscillating liquid column in a U-tube is

(a) Periodic but not simple harmonic.

(b) Non-periodic.

(c) Simple harmonic and time period is independent of the density of the liquid.

(d) Simple harmonic and time-period is directly proportional to the density of the liquid.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) Consider the diagram in which a liquid column oscillates . in this case, restoring force acts on the liquid due to gravity.

Restoring force f = weight of liquid column of height 2y

t=-A $* 2 y * ρ * g$ = -2A $ρ g y$

f $\propto - y$

motion is SHM with force constant k= 2A $ρ g$

T= $2 π \sqrt[]{\frac{m}{k}}$ = 2 $π \sqrt[]{\frac{A * 2 h * ρ}{2 A ρ g}} = 2 π \sqrt[]{\frac{h}{g}}$

So time period is independent upon density of liquid.

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

Motion of an oscillating liquid column in a U-tube is

(a) Periodic but not simple harmonic.

(b) Non-periodic.

(c) Simple harmonic and time period is independent of the density of the liquid.

(d) Simple harmonic and time-period is directly proportional to the density of the liquid.

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

The relation between acceleration and displacement of four particles are given below:

(a) a_x = + 2x.

(b) a_x = + 2x².

(c) a_x = – 2x².

(d) a_x = – 2x. Which one of the particles is executing simple harmonic motion?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) For motion to be in SHM acceleration of the particle must be proportional to negative of displacement.

a $\propto - y$ , so y has to linear.

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

The displacement of a particle is represented by the equation 3 y t = sin ω . The motion is

(a) Non-periodic.

(b) Periodic but not simple harmonic.

(c) Simple harmonic with period 2π/ω.

(d) Simple harmonic with period π/ω.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) y = sin³wt

= [3sinwt-4sin3wt]/4

dy/dt= [ $\frac{d}{d t} (3 s i n w t) - \frac{d}{d t} (4 s i n 3 w t)$ ]/4

4dy/dt=3wcoswt-4 [3wcoswt]

4 $* \frac{d^{2} y}{d t^{2}} = - 3 w^{2} s i n w t + 12 w s i n 3 w t$

$\frac{d^{2} y}{d t^{2}}$ = - $\frac{3 w^{2} s i n w t + 12 w^{2} s i n 3 w t}{4}$

$\frac{d^{2} y}{d t^{2}}$ is not proportional to y. hence it is not SHM.

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

The displacement of a particle is represented by the equation y = 3cos( $\frac{π}{4} - 2 w t$ ). The motion of the particle is

(a) Simple harmonic with period 2 $π$ /w.

(b) Simple harmonic with period π/ω.

(c) Periodic but not simple harmonic.

(d) Non-periodic.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Velocity =dy/dt= $\frac{d}{d t} [3 c o s (\frac{π}{4} - 2 w t)]$

= 3 (-2w) [-sin ( $\frac{π}{4} - 2 w t$ )]

= 6wsin $[\frac{π}{4} - 2 w t]$

Acceleration a = dv/dt= $\frac{d}{d t} [6 w s i n (\frac{π}{4} - 2 w t)]$

= -4w² [3cos ( $\frac{π}{4} - 2 w t$ )]

A = -4w²y hence acceleration is directly proportional to displacement so it follows SHM

w'= 2w

2 $\frac{π}{T} = 2 w$

T'= 2 $π / 2 w$ = $\frac{π}{w}$

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Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen Class 11th

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

By considering the diagram

_{$θ$
1}= $θ_{o} s i n (w t + \emptyset_{1})$

_{$θ$
2}= $θ_{o} s i n ? (w t + \emptyset_{2})$

As it is clear given that amplitude time period being equal but phases being different. Now for first pendulum at any time t

_{$θ$
1}= $θ$ ₂

So sin $\frac{π}{2}$ = sin(wt+ $\emptyset_{1}$ )

wt+ $\emptyset_{1} = \frac{π}{2}$

where $θ$ _o=2^o is the angular amplitude of first pendulum . for the second pendulum , the angular displacement is one degree , therefore $θ$ ₂= $\frac{θ_{0}}{2}$ and negative sign is taken to show for being left to mean position.

- $\frac{θ_{o}}{2} = θ_{0} s i n (w t + θ_{2})$

Sin(wt+ $\emptyset_{2}$ )=-1/2

So (wt+ $θ$ ₂)=- $\frac{π}{6} o r \frac{7 π}{6}$

So by making their difference

(wt+ $θ$ ₂)-( $w t + θ$ ₁)=7 $\frac{π}{6} - \frac{π}{2}$ =4 $\frac{π}{6}$

( $θ$ _{2-
$θ$} ₁)= 120⁰

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