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6 months ago

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

A body is performing S.H.M. Then its

(a) Average total energy per cycle is equal to its maximum kinetic energy.

(b) Average kinetic energy per cycle is equal to half of its maximum kinetic energy.

(c) Mean velocity over a complete cycle is equal to 2 π times of its maximum velocity.

(d) Root mean square velocity is 1/√ 2 times of its maximum velocity

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

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

E= $\frac{1}{2} m$ w²a²

KE=1/2mv²= $\frac{1}{2} m [{\frac{d x}{d t}]}^{2}$

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= $\sqrt[]{\frac{v_{1}^{2} + v_{2}^{2}}{2}} = \sqrt[]{\frac{0 + a^{2} w^{2}}{2}} = \frac{a w}{\sqrt[]{2}}$

V_rms=V_max/ $\sqrt[]{2}$

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6 months ago

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

The displacement time graph of a particle executing S.H.M. is shown in Fig. 14.5. Which of the following statement is/are true?

(a) The force is zero at t =3T/4 .

(b) The acceleration is maximum at t = 4T/4.

(c) The velocity is maximum at t = T/4.

(d) The P.E. is equal to K.E. of oscillation at t = T/2 .

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Payal Gupta

Contributor-Level 10

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.

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6 months ago

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

Which of the following statements is/are true for a simple harmonic oscillator?

(a) Force acting is directly proportional to displacement from the mean position and opposite to it

(b) Motion is periodic.

(c) Acceleration of the oscillator is constant.

(d) The velocity is periodic.

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Payal Gupta

Contributor-Level 10

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 $* a c c e l e r a t i o n$ = -mw²x

So force is directly proportional to displacement.

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6 months ago

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

Displacement vs. time curve for a particle executing S.H.M. is shown in Fig. 14.4. Choose the correct statements. Fig. 14.4 Fig. 14.3

(a) Phase of the oscillator is same at t = 0 s and t = 2 s.

(b) Phase of the oscillator is same at t = 2 s and t = 6 s.

(c) Phase of the oscillator is same at t = 1 s and t = 7 s.

(d) Phase of the oscillator is same at t = 1 s and t = 5 s.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) It is clear the curve that points corresponding to t=2s and t=6s are separated by a distance belongs to one time period. hence these points must be in same phase

Similarly points belongs t-1s to t=5sare at separation of one time period. Hence must be in phase.

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6 months ago

Class 11th 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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6 months ago

Class 11th 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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6 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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6 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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6 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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6 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