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Mere Teacher · Previous Year Questions
NEET Physics · Simple Harmonic Motion · Set 5 (Q121-150)
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121.
A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as:
(1)T=2πv12−v22x22+x12
(2)T=2πv12+v22x22+x12
(3)T=2πv12+v22x22−x12
(4)T=2πv12−v22x22−x12
122.
A body executes simple harmonic motion with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy?
(1)A/4
(2)A/2
(3)3A/4
(4)Some other fraction of A
123.
A particle performs S.H.M. with amplitude 25cm and period 3s. The minimum time required for it to move between two points 12.5cm on either side of the mean position is
(1)0.6s
(2)0.5s
(3)0.4s
(4)0.2s
124.
The potential energy of a simple harmonic oscillator of mass 2kg at its mean position is 5J. If its total energy is 9J and amplitude is 1cm, then its time period is
(1)100π s
(2)50πs
(3)20πs
(4)10πs
125.
A particle executes simple harmonic motion with a time period 0.6s and amplitude 10cm. Then the mean velocity of the particle over the time interval during which it travels a distance 5cm starting from the equilibrium position.
(1)1ms−1
(2)50cms−1.
(3)10cms−1
(4)1cms−1
126.
When a body is suspended from a fixed point by a spring, the angular frequency of its vertical oscillations is ꞷ1. When a different spring is used, the angular frequency is ꞷ2. The angular frequency of vertical oscillations when both the springs are used together in series is given by
(1)ω=(ω12+ω22)21
(2)ω=(2ω12+ω22)21
(3)ω=((ω12+ω22)ω12ω22)21
(4)ω=(2(ω12+ω22)ω12ω22)21
127.
A mass falls from a height ' h ' and its time of fall 't' is recorded in terms of time period T of a simple pendulum. On the surface of earth, it is found that t=2T. The entire set up is taken on the surface of another planet whose mass is half of earth and radius the same. Same experiment is repeated and corresponding times noted as t’ and T’. [NEET 2019]
(1)t’=2T’
(2)t’>2T’
(3)t’<2T’
(4)t’=2T’
128.
The equation of a particle executing simple harmonic motion is given by x=sinπ(t+31)m. At t=1s, the speed of particle will be (Given: π=3.14)
(1)0cms−1
(2)157cms−1
(3)272cms−1
(4)314cms−1
129.
Time period of a simple pendulum is 2 sec. If its length is increased by 4 times, then its period becomes
(1)8sec
(2)12sec
(3)16 sec
(4)4 sec
130.
The mechanical energy of a damped oscillator becomes half of its initial energy in 4seconds. In another t seconds its mechanical energy becomes 12.5% of its initial mechanical energy. Then t=
(1)4s
(2)8s
(3)12s
(4)16s
131.
The ratio of frequencies of two oscillating pendulums are 3:2. Their lengths are in the ratio
(1)3:2
(2)9:4
(3)4:9
(4)2:3
132.
A particle rests in equilibrium under two forces of repulsion whose centers are at distance of a and b from the particle. The forces vary as the cube of the distance. The forces per unit mass are k and k' respectively. If the particle be slightly displaced towards one of them the motion is simple harmonic with the time period equal to
(1)3(a3k+b3k’)2π
(2)(a3k+b3k’)2π
(3)(a4k+b4k’)2π
(4)3(ak+bk’)2π
133.
Given below are two statements: Statement (I):- A second's pendulum has a time period of 1 second. Statement (II):- It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options give below.
(1)Both Statement I and Statement II are false
(2)Statement I is true but Statement II is false
(3)Statement I is false but Statement II is true
(4)Both Statement I and Statement II is true
134.
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be
(1)α2β2
(2)βα
(3)αβ2
(4)α2πβ
135.
A particle moves along the X-axis according to the equation x = 10 sin3 (πt). The amplitudes and frequencies of component SHMs are
(1)amplitude 30/4, 10/4 ; frequencies 3/2, ½
(2)amplitude 30/4, 10/4 ; frequencies 1/2, 3/2
(3)amplitude 10, 10 ; frequencies 1/2, ½
(4)amplitude 30/4, 10 ; frequencies 3/2, 2
136.
For a periodic motion represented by the equation y=sinωt+cosωt the amplitude of the motion is
(1)1
(2)0.5
(3)2
(4)2
137.
particle performing S.H.M. starts from equilibrium position and its time period is 16 second. After 2 seconds its velocity is πm/s. Amplitude of oscillation is (cos45∘=21)
(1)22m
(2)42m
(3)62m
(4)82m
138.
A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then:
(1)amplitude of motion is 3a
(2)time period of oscillations is 8τ
(3)amplitude of motion is 4a
(4)time period of oscillations is 6τ
139.
A spring of force constant k is cut into lengths of ratio 1:2: 3. They are connected in series and the new force constant is k’. Then they are connected in parallel and force constant is k Then k’ : k″ is
(1)1:9
(2)1: 11
(3)1:14
(4)1:6
140.
A particle of mass 0.1kg is executing simple harmonic motion of amplitude 0.1m. When the particle passes through the mean position, its kinetic energy is 8×10−3J. If the initial phase is 45∘, the equation of its motion is (Assume, x(t) as the position of the particle at time t.)
(1)x(t)=0.1sin(4t+4π)
(2)x(t)=0.1sin(16t+4π)
(3)x(t)=0.1sin(2(t+4π))
(4)x(t)=0.1sin(2t+4π)
141.
Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3 s and 4 s respectively. If their moments of inertia are in the ratio of 3 : 2 then the ratio of their magnetic moments will be:
(1)2:1
(2)8:3
(3)1 : 3
(4)27 : 1
142.
A particle executing SHM has maximum speed of 0.5ms−1 and maximum acceleration of 1.0ms−2. The angular frequency of oscillation is
(1)2rads−1
(2)0.5rads−1
(3)2πrads−1
(4)0.5πrads−1
143.
Under the action of a force F=−75y where F is in Newton and y is in meters, an object of mass 3kg executes simple harmonic motion. If the velocity of the object at the mean position is 2.5ms−1, the maximum acceleration of the object is
(1)5ms−2
(2)7.5ms−2
(3)10ms−2
(4)12.5ms−2
144.
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is [NEET 2020]
(1)23πrad
(2)2πrad
(3)zero
(4)πrad
145.
The transverse displacement of a string of a linear density 0.01kgm−1, clamped at its ends is given by Y(x,t)=0.03sin(32πx)cos(60πt), where x and y are in metres and time t is in seconds. Tension in the string is
(1)9N
(2)36N
(3)162N
(4)81N
146.
If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0s to t = τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:
(1)b0.693
(2)b
(3)b1
(4)b2
147.
Two particles execute simple harmonic motion along the same straight line with same amplitude and 'same frequency. The two particles pass one another when moving in opposite directions each time at a distance of 21 times the amplitude from their common mean position. The phase difference between the two particles is
(1)30∘
(2)45∘
(3)60∘
(4)90∘
148.
A second’s pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is
(1)moving down with uniform acceleration
(2)moving around the earth in geostationary orbit
(3)moving up with uniform velocity
(4)moving up with uniform acceleration.
149.
A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is k’. If they are connected in parallel and force constant is k’’, then k’:k’’ is [NEET 2017]
(1)1:6
(2)1:9
(3)1:11
(4)1:14
150.
A simple harmonic oscillator of frequency 1Hz has a phase of 1rad. By how much should the origin be shifted in time, so as to make the phase of the oscillator vanish? [Time in seconds (s)].