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Mere Teacher · Previous Year Questions
NEET Physics · Simple Harmonic Motion · Set 11 (Q301-328)
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301.
If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 0.02 m/s2 at any time, the angular frequency of the oscillator is equal to Oscillations 123
(1)10 rad/s
(2)0.1rad/s
(3)100 rad/s
(4)1 rad/s
302.
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is:
(1)π2ma2v2
(2)41ma2v2
(3)4π2ma2v2
(4)2π2ma2v2
303.
Amplitude of a wave is represented by A=a+b−cc Then resonance will occur when
(1)b=−c/2
(2)b = 0 and a = c
(3)b=−a/2
(4)None of these
304.
A mass is suspended from a vertical spring which is executing. S.H.M. of frequency 5Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is [acceleration due to gravity g=10m/s2 ]
(1)2πm/s
(2)πm/s
(3)2π1m/s
(4)π1 m/s
305.
A body ofmass5kg hangs from a spring and oscillates with a time period of 2π seconds. If the ball is removed, the length of the spring will decrease by
(1)kg metres
(2)gk metres
(3)2π metres
(4)g metres.
306.
Consider two SHMs along the same straight line x1=A1sin(ωt+φ1),x2=A2sin(ωt+φ2), where A1 and A2 are their amplitudes and φ1 and φ2 are their initial phase angle. If the two SHMs meet simultaneously and ' R ' is the resultant amplitude, match column I with column II. Column-I(A).The two SHMs are in phase, A1=A2=A(B).The two SHMs are in phase, A1=A2(C).The two SHMs are 90 out of phase, A1=A2=A(D).The two SHMs are 180∘ out of phase, A1=A2Column-II(I).R=A1+A2(II).R=0(III).R=2A(IV).R=2A
(1)A-III, B-I, C-IV, D-II
(2)A-IV, B-III, C-II, D-I
(3)A-I, B-III, C-II, D-IV
(4)A-III, B-IV, C-I, D-II
307.
Resonance is an example of
(1)Tuning fork
(2)Forced vibration
(3)Free vibration
(4)Damped vibration
308.
A simple pendulum of length 2m is given a horizontal push through angular displacement of 60∘. If the mass of bob is 200gram, the angular velocity of the bob will be (Take acceleration due to gravity =s210m ) (sin30∘=cos60∘=0.5,cos30∘=sin60∘=23)
(1)22rad/s
(2)32rad/s
(3)22.5rad/s
(4)32.5rad/s
309.
When two displacements represented by y1=asin(ωt) and y2=bcos(ωt) are superimposed the motion is
(1)simple harmonic with amplitude a2+b2
(2)simple harmonic with amplitude 2(a+b)
(3)not a simple harmonic
(4)simple harmonic with amplitude ba
310.
The displacement of a particle varies according to the relation x = 4(cosπt + sinπt). The amplitude of the particle is
(1)8
(2)– 4
(3)4
(4)42
311.
Which of the following statements regarding damping force of a damped oscillator is NOT correct?
(1)Damping force depends on the nature of the surrounding medium.
(2)Damping force is generally proportional to the velocity of the body making oscillations.
(3)Damping force acts in the direction of the velocity of the body.
(4)Ratio of the damping force and velocity of the body depends on the size and shape of the body.
312.
Two masses m and 2m are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system (see figure). Because of torsional constant k, the restoring toruque is τ=kθ for angular displacement θ. If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position will be
(1)l3kθ02
(2)l2kθ02
(3)lkθ02
(4)2lkθ02
313.
Two blocks of masses m and M,(M>m), are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released, then ( μ= coefficient of friction between the two blocks)(A) The time period of small oscillation of the two blocks is T=2πk(m+M)(B) The acceleration of the blocks is a=−M+mkx ( x= displacement of the blocks from the mean position) (C) The magnitude of the frictional force on the upper block is M+mmμ|x|(D) The maximum amplitude of the upper block, if it does not slip, is kμ(M+m)g(E) Maximum frictional force can be μ(M+m)g. Choose the correct answer from the options given below:
(1)B, C, D Only
(2)C, D, E Only
(3)A, B, D Only
(4)A, B, C Only
314.
A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency ω=3πrad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45º as shown in the figure. Point P is at a horizontal distance of 10 cm from O. If the pebble hits the block at t = 1s, the value of v is (take g = 10 m/s2)
(1)50m/s
(2)51m/s
(3)52m/s
(4)53m/s
315.
In case of a forced vibration, the resonance peak becomes very sharp when the
(1)damping force is small
(2)restoring force is small
(3)applied periodic force is small
(4)quality factor is small
316.
An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D,E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:
(1)I=1615I0
(2)I=43I0
(3)I=169I0
(4)I=4I0
317.
The total energy of particle performing SHM depend on
(1)k,a,m
(2)k,a
(3)k,a,x
(4)k,x.
318.
The S.H.M. of a particle is given by the equation y = 3 sin ωt + 4 cos ωt. The amplitude is:
(1)7
(2)1
(3)5
(4)12
319.
The S.H.M. of a particle is given by the equation y=3sinωt+4cosωt. The amplitude is
(1)7
(2)1
(3)5
(4)12
320.
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.
The potential energy U(x)versus time (t)plot of the particle is correctly shown in figure:
(1)
(2)
(3)
(4)
321.
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is:
(1)π5
(2)2π5
(3)54π
(4)32π
322.
The amplitude of a particle due to superposition of following S.H.Ms. Along the same line is X1 = 2 sin 50πt ; X2 = 10 sin (50 πt + 37º) X3 = - 4 sin 50πt ; X4 = - 12 cos 50πt
(1)4 2
(2)4
(3)6 2
(4)none of these
323.
When two displacement represented by y1 = a sin (ꞷt) and y2 = b cos (ꞷt) are superimposed the motion is
(1)simple harmonic with amplitude ba
(2)simple harmonic with amplitude a2+b2
(3)simple harmonic with amplitude 2(a+b)
(4)not a simple harmonic
324.
A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to:
(1)Air friction
(2)Moment of inertia
(3)Weight of the bob
(4)Combination of all the above
325.
The resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by 2π is
(1)simple harmonic
(2)circular
(3)elliptical
(4)parabolic
326.
A block is in simple harmonic motion (S.H.M) on the end of the spring with position given by x= (5) cos(ωt+4π)cm. If the total mechanical energy used is 100J to achieve maximum displacement, then the potential energy at time t=0 is
(1)20J
(2)80J
(3)75J
(4)50J
327.
A block of mass 100g is connected to an elastic spring of spring constant 450Nm−1 oscillates vertically. The block-spring system is in viscous surrounding medium with a damping constant 69.3gs−1. The time in which the amplitude of oscillations drop to half of its initial value. [take ln2=0.693 ]
(1)6.93s
(2)2s
(3)20s
(4)69.3s
328.
A simple pendulum oscillates in air with time period T and amplitude A. As the time passes