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4 months ago

Oscillations Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is

(a) Simple harmonic motion.

(b) Non-periodic motion.

(c) Periodic motion.

(d) Periodic but not S.H.M.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) consider the motion of a ball inside a smooth curved bowl

for small angular displacement or slightly released motion, it can be considered as angular SHM.

When the ball is at angle $θ$ of the restoring force acts

Ma=mgsin $θ$

A= gsin $θ$

$\frac{d^{2} x}{{d t}^{2}} = - g s i n θ = - g \frac{x}{R}$

Hence motion is SHM

So T= $\frac{2 π}{w} = 2 π \sqrt[]{\frac{R}{g}}$

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4 months ago

Oscillations Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen

The rotation of earth about its axis is

(a) Periodic motion.

(b) Simple harmonic motion.

(c) Periodic but not simple harmonic motion.

(d) Non-periodic motion.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) the rotation of earth about its axis is periodic because it repeats after a regular interval of time.

The rotation of earth is obviously not a to and fro type of motion about a fixed point. So its motion is SHM.

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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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4 months ago

Oscillations 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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