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5 months ago

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

When a mass m is connected individually to two springs S₁ and S₂, the oscillation frequencies are ν₁ and ν₂ . If the same mass is attached to the two springs as shown in Fig. 14.3, the oscillation frequency would be

(a) ν₁ + ν₂

(b) $\sqrt[]{v_{1}^{2} + v_{2}^{2}}$

(c) ( $\frac{1}{v_{1}} + \frac{1}{v_{2}})$ ^-1

(d) $\sqrt[]{v_{1}^{2} - v_{2}^{2}}$

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) K_eq= K₁+K₂

Time period of oscillation of the spring block system

T=2 $π \sqrt[]{\frac{m}{K_{e q}}} = 2 π \sqrt[]{\frac{m}{K_{1} + K_{2}}}$

Frequency = 1/time = $\frac{1}{2 π} \sqrt[]{\frac{K_{1} + K_{2}}{m}}$ = equivalent oscillation frequency

When the mass is connected to the springs individually

v_{1} = \frac{1}{2 π} \sqrt[]{\frac{k_{1}}{m}}

v_{2} = \frac{1}{2 π} \sqrt[]{\frac{k_{2}}{m}}

From above equation

$v = \frac{1}{2 π} {[\frac{k_{1}}{m} + \frac{k_{2}}{m}]}^{\frac{1}{2}}$

$v = \frac{1}{2 π} {[\frac{4 π^{2} v_{1}^{2}}{1} + \frac{4 π^{2} v_{2}^{2}}{1}]}^{\frac{1}{2}}$

So $v = \sqrt[]{v_{1}^{2} + v_{2}^{2}}$

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5 months ago

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

A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². The period of oscillation is

(a) π s.

(b) π/2 s.

(c) 2π s.

(d) π/t s.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) V=dy/dt=awcoswt

V_max=aw=30

Acceleration A= $\frac{d x^{2}}{d t^{2}} = - a w^{2} s i n w t$

A=w²a=60

So after solving w =2rad/s

2 $\frac{π}{T} = 2 r a d / s$

T= $π s e c$

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5 months ago

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

The equation of motion of a particle is x = a cos (α t ) 2. The motion is

(a) Periodic but not oscillatory.

(b) Periodic and oscillatory.

(c) Oscillatory but not periodic.

(d) Neither periodic nor oscillatory.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

X (t+T)=acos [ $α (t + T)$ ]²

= acos [ $α t^{2} + α T^{2} + 2 α t T$ ]

So it is not periodic.

So it is oscillatory but not periodic

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5 months ago

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

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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5 months ago

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

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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5 months ago

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

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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5 months ago

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

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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5 months ago

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

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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New question posted

5 months ago

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

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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New answer posted

5 months ago

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

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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