Test: Oscillations 1 - From Past 28 Years Questions


25 Questions MCQ Test Physics 28 Years Past year papers for NEET/AIPMT Class 11 | Test: Oscillations 1 - From Past 28 Years Questions


Description
This mock test of Test: Oscillations 1 - From Past 28 Years Questions for NEET helps you for every NEET entrance exam. This contains 25 Multiple Choice Questions for NEET Test: Oscillations 1 - From Past 28 Years Questions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Oscillations 1 - From Past 28 Years Questions quiz give you a good mix of easy questions and tough questions. NEET students definitely take this Test: Oscillations 1 - From Past 28 Years Questions exercise for a better result in the exam. You can find other Test: Oscillations 1 - From Past 28 Years Questions extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph. [2003]

Solution:

Potential energy ∝ (Amplitude)2 ∝ (Displacement)2 P.E.is  maximum at maximum distance.
P.E. is zero at equilibrium point.
P.E. curve is parabolic.

QUESTION: 2

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is [2003]

where E is the total energy

Solution:

At half way

QUESTION: 3

In case of a forced vibration, the resonance wave becomes very sharp when the [2003]

Solution:

The resonan ce wave becomes very sharp when damping force is small.

QUESTION: 4

The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

Solution:

When a spring is cut into n parts Spring constant for each part = nk Here, n = 4

QUESTION: 5

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion ?

Solution:

When v is maximum, a is zero

QUESTION: 6

Two spr ings of spring constants k1 and k2 are joined in series. The effective spring constant of the combin ation is given by [2004]

Solution:

QUESTION: 7

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is

Solution:

In S.H.M. , 

QUESTION: 8

The potential energy of a long spring when stretched by 2cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is

Solution:

The potential energy of a spring

For x = 8 cm,

Energy stored

QUESTION: 9

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is [2007]

Solution:

Let y = A sinωt

QUESTION: 10

The particle executing simple harmonic motion has a kinetic energy K0 cos2ωt . The maximum values of the potential energy and the total energy are respectively [2007]

Solution:

We have, U + K = E where, U = potential energy, K = Kinetic energy, E = Total energy.
Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and viceversa.

Further,

But by question, 2 K .E. = K0 cosωt

Hence, total energy, 

∴  Umax = K0 & E = K0.

QUESTION: 11

A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take g = 10 m/s2)?[2007]

Solution:

Mass gets detached at the upper extreme position when pan returns to its mean position.
R = mg – mω2a = 0

⇒ a = 1/10m = 10cm

QUESTION: 12

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is[2007]

Solution:

Displacement from the mean position

According to problem y = a/2

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

QUESTION: 13

Two simple harmonic motions of angular frequency 100 and 1000 rad s–1 have the same displacement amplitude. The ratio of their maximum accelerations is: [2008]

Solution:

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e.

QUESTION: 14

A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Solution:

Maximum velocity = aω

According to question,

QUESTION: 15

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at will be:[2009]

Solution:

Speed

QUESTION: 16

Which one of the following equations of motion represents simple harmonic motion? [2009]

where k, k0, k1 and a are all postive.

Solution:

a = – kX, X = x + a.
In simple harmonic motion acceleration is directly proportional to the displacement from the mean position. Also the acceleration is in the opposite direction of displacement.

QUESTION: 17

A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be: [2009]

Solution:

If x is the maximum extension produced, we have
Mgx=1/2kx2

, on equating the decrease in the gravitational potential energy of the mass M to the increase in the elastic potential energy of the spring.

Therefore, x = 2Mg/k

QUESTION: 18

The displacement of aparticle along the x-axis is given by x = a sin2ωt. The motion of the particle corresponds to: [2010]

Solution:

This represents an S. H. M. of frequency

QUESTION: 19

The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be [2010]

Solution:

QUESTION: 20

A par ticle of mass m is r eleased fr om rest and follows a parabolic path as shown.
Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time [2011]

Solution:

The given velocity-position graph depicts that the motion of the particle is SHM.
In SHM, at t = 0, v = 0 and x = xmax
So, option (a) is correct.

QUESTION: 21

 Out of the following functions, representing motion of a particle, which represents SHM?
(A) y = sin ωt - cosωt [2011]
(B) y = sin3 ωt

(C) 

(D) y = 1 + ωt + ω2t2

Solution:

Only functions given in (A) & (C) represent SHM.

QUESTION: 22

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is [2011M]

Solution:

Equation of SHM is given by x = A sin (ωt + δ) (ωt + δ) is called phase.

For second particle,

QUESTION: 23

The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are : [2012]

Solution:

QUESTION: 24

The equation of a simple harmonic wave is given by

Where x and y are in meters and t is in seconds.The ratio of maximum particle velocity to the wave velocity is [2012M]

Solution:

on comparing with the standard wave equation

y = a sin (ωt – kx) Wave velocity

The velocity of particle

QUESTION: 25

A particle of mass m oscillates along x-axis according to equation x = a sinωt. The nature of the graph between momentum and displacement of the particle is [NEET Kar. 2013]

Solution:

This is the equation ofellipse. Hence the graph is an ellipse. P versus x graph is similar to V versus x graph.