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Physics NEET MCQ
Oscillations And Wave Mcq
Quiz 4
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Q.1
The resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by π/2 is [ MPPMT 1989]
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a) simple harmonic
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b) circular
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c)elliptical
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d)parabolic
Explanation
Answer: (c)
Q.2
Due to some force F1 a body oscillates with period 4/5 sec and due to other force F2 it oscillates with period 3/5 sec. If both force acts simultaneously new period will be [ Raj.PMT 1997]
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a) 0.72 s
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b) 0.64 s
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c)0.68s
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d)0.36s
Explanation
y=asin(2πt / T) for force F1 y1=a sin( 2π×5t / 4) and for force F2 y2=a sin(2π×5t / 3) Now y=y1+y2y=2sin(35πt/12 ) cos(5πt/12) from above 2πt/T=35πt/12=0.68s Answer:(c)
Q.3
Two simple harmonic motions of same amplitude, same frequency and a phase difference of π/4 are superimposed at right angles to each other on a particle, the particle will describe a ..[ MPPMT 1990]
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a) circle
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b) ellipse
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c) figure of eight
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d) straight
Explanation
Answer: (b)
Q.4
The motion of a particle is expressed by the equation a=-bX, where a is the acceleration, X is the displacement from the equilibrium position and b is a constant. the periodic time will be [ MNR 1995]
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a)2π/b
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b) 2π/ √b
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c) 2π√b
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d)2√(π/b)
Explanation
Comparing with differential equation for SHM acceleration a=-ω2x, we get ω2=b (2π/T)2=b T=2π/√bAnswer: (b)
Q.5
A mass M is suspended from a spring of negligible mass. the spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, then the time period becomes (5/4)T. The ratio (m/M) is [ CPMT 1991]
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a) 9/16
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b) 25/16
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c)4/5
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d)5/4
Explanation
Answer: (a)
Q.6
Two mutually perpendicular simple harmonic vibrations have same amplitude, frequency and phase. when they superimposed, the resultant form of vibration will be
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a) a circle
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b) an ellipse
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c)a straight line
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d)a parabola
Explanation
Answer:(c)
Q.7
For a simple pendulum the graph between L and T will be ..[ CPMT 1992]
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a) hyperbola
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b) parabola
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c) a curver line
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d) a straight line
Explanation
Answer: (b)
Q.8
A simple pendulum is set up on a trolley which moves to the right with an acceleration a on a horizontal plane. then the threads of the pendulum in the mean position makes an angle of θ with the vertical
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a) tan-1 (a/g) in the forward direction
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b) tan-1 (a/g) in the backward direction
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c) tan-1 (g/a) in the backward direction
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d) tan-1 (g/a) in the forward direction
Explanation
As shown in figure the reaction force on the pendulum produces a backward horizontal acceleration a in the pendulum ∴ tanθ = a/g Answer: (b)
Q.9
When a particle oscillates simple harmonically, its kinetic energy vary periodically. If frequency of the particle is n, the frequency of the kinetic energy is .. [ MNR 1994]
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a) 4n
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b) n
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c) 2n
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d) n/2
Explanation
Answer: (c)
Q.10
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, its displacement from equilibrium in terms of its amplitude 'a' is [ MPPMT 1995]
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a) a/4
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b) a/3
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c) a/2
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d) 2a/3
Explanation
Maximum potential energy = ½ mω2a2 Given (1/4)½ mω2a2 = ½ mω2y2 y = a/2 Answer:(c)
Q.11
A particle is executing SHM of amplitude a with a time-period T sec. The time taken by it to move from positive extreme position to half of the amplitude is
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a) T/12 sec
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b) 2T/12 sec
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c) 3T/12 sec
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d) 6T/12 sec
Explanation
y = asinωt Thus a = a sin ωt1 ωt1 = π/2 t1 = π/2ω = T/4 sec and a/2 = asinωt2 ωt2 = π/6 t2 = T/6 Δt = T/4 -T/12 = 2T/12 Answer: (b)
Q.12
A cylindrical piston of mass M slides smoothly inside a long cylinder at one end enclosing a certain mass of gas. the cylinder is kept with its axis horizontal. if the piston is disturbed from its equilibrium position, it oscillates simple harmonically. the period of oscillation will be
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a)
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b)
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c)
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d)
Explanation
When piston is moved inside then volume will decrease and pressure will increaseLet P1=P=Initial pressureV1=V=Initial volume let P2=P + ΔP=Final pressureV2=V - ΔV=Final Volume Using Boil's law P1V1=P2V2ORPV=(P+ΔP ) (V-ΔV) PV=PV + VΔP - PΔV - ΔPΔV Since ΔV and ΔP is very small VΔP=PΔV ΔP=PΔV /V Multiplying above equation by A AΔP=(PAΔV) /V Restoring force acting on he piston opposite to displacement due to excess pressure F=AΔP F=(PAΔV) /V ΔV=Ax here x is displacement of piston F=(PA2x) /V F=(PA2 /V) x comparing above equation with F=kx equation for SHM we get k=PA2 /V Now period T=2π(m/k) 2 Answer: (a)
Q.13
Figure below shows a mass m suspended with a mass less inextensible string passing over a frictionless pulley. The spring constant K. The time period of oscillation of mass m is
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a) 2π √(m/2k)
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b) 2π √(m/k)
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c)2π √(2m/k)
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d)none of the above
Explanation
mg=ky acceleration g=(k/m) y Standard equation for acceleration a=-ω2 y ∴ ω=√(k/m)2π/T=√(k/m)T=2π /√(k/m) ∴ T=2π √(m/k)Answer: (b)
Q.14
A mass m is attached to a massless inextensible string as shown in figure. the string passes over a frictionless pulley which in turn is connected to spring. If the mass is displaced from its position then the time period of oscillation of the mass m is
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a) 2π√(m/2k)
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b) 2π√(m/k)
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c)4π√(m/k)
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d)π√(m/k)
Explanation
Force stretching the spring=2T=2mg. If the mass moves down by y, then the extension in the spring=y/2∴ 2mg=k (y/2) acceleration=g=(k/4g)y ∴ ω=(k/4g)T=4π √(m/k) Answer:(c)
Q.15
The displacement y in cms is given in terms of time t second by the equation y = 3sin314t + 4cos314t then the amplitude of SHM is [ MPPET 1993]
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a) 7 cm
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b) 3 cm
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c) 4 cm
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d) 5 cm
Explanation
Let 3 = A cosθ and 4 = Asinθ Thus 32 + 42 = A2 A = 5 cm Answer: (d)
Q.16
A tunnel is made across the earth passing through the centre. A ball is dropped from a height h in the tunnel. The motion will be periodic with time period
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a)
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b)
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c)
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d)
Explanation
the time taken by the ball to reach the surface of the earth = √(2h/g). the time period of oscillation in the tunnel = 2π√(R/g) and then ball moves to other side to a height h and takes time 2√(2h/g) to go back there and return Then time time taken to reach height h on the other side ( i.e. from where the ball is dropped ) = √(2h/g) ∴ periodic time Answer: (c)
Q.17
A mass M is suspended from a light spring. An additional mass m added displaces the spring further by distance x. Now the combined mass will oscillate on the spring with period [ CPMT 1989]
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a)
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b)
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c)
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d)
Explanation
mg = kx k = mg/x Periodic time T = 2π√( mass/ k) mass = M+m Answer:(b)
Q.18
in figure S1 and S2 are identical springs. The oscillation frequency of the mass m is f. if one spring is removed, the frequency will become..
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a) f
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b) 2f
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c) √2 f
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d) f/ √2
Explanation
As shown in figure both the spring are connected in parallel. If k is the restoring constant of the combination then after removal of one spring new restoring constant become k' = k/2 frequency of oscillation = (1/2π) √(k/m) when both the spring are connected f' = (1/2π) √(k'/m) f' = (1/2π) √(k/2m) f' =f / √2 Answer: (d)
Q.19
A simple pendulum consisting of a ball of mass m tied to a string of length l is made to swing on a circular arc of an angle θ in a vertical plane. At the end of this arc, another ball of mass m is placed at rest. the momentum transferred to this ball at rest by the swinging ball is
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a)zero
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b) mθ / √(l/g)
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c)mθl / √(l/g)
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d)ml2π√(l/g)
Explanation
Velocity of the bob is zero at highest point of the path. it has no momentum itselfAnswer: (a)
Q.20
A simple pendulum performing simple harmonic motion about X=0 with the amplitude A and time period T. the speed of the pendulum at X=A/2 will be [ MPPMT 1987]
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a) πA√3 /T
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b) πA /T
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c)πA√3 /2T
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d)3π2A / T
Explanation
Formula for velocity Answer: (a)
Q.21
For a particle executing simple harmonic motion the kinetic energy K is given by K=Kocos2ωt. The maximum potential energy is
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a) Ko
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b) zero
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c)Ko/2
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d)not obtainable
Explanation
maximum potential energy=maximum Kinetic energy for maximum kinetic energy cosωt=1 ∴ Maximum potential energy=Ko Answer:(a)
Q.22
A body executing S.H.M when its displacement from the mean position is 4cm and 5cm, the corresponding velocity of the body is 10cm per sec and 8cm per sec. then the time period of the body is [ CPMT 1991]
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a) 2π sec
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b) (π/2) sec
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c) π sec
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d) (3π/2) sec
Explanation
Form the formula for velocity Answer: (c)
Q.23
Om a smooth inclined plane a body of mass M is attached between two springs. the other ends of the springs on fixed to firm supports. If each spring has force constant k, the period of oscillation of the body ( assume the spring are massless) is ...
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a)
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b)
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c)
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d)
Explanation
Period of oscillation of spring is T=2π √(M/k)Both the springs are connected in parallel thus k'=2k ∴ T'=2π √(M/2k) note slope is irrelevant as g does not affect the motionAnswer: (a)
Q.24
The time period of pendulum of infinite length is ..[ RajPET 1996]
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a) Infinity
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b) Zero
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c)84.6 min
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d)Nothing certain
Explanation
Time period of a very big pendulum having length comparable to that of radius of earth isfor initiate length tale l=∞Answer: (c)
Q.25
Two identical springs of force constant 'k' are connected in a) series b) in parallel. the combinations support mass 'm' at lower end. The ratio of the period of oscillations mass 'm' in series and parallel combination is ..[ MPPMT 1993]
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a) 1:1
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b) 1:2
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c)1:4
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d)2:1
Explanation
When connected in series k'=k/2Thus Period T' ∝ √ (2/k)When connected in parallel k"=2kThus period T"∝ √(1/2k)T'/T"=√(2/k)/√(1/2k)T'/T"=√(4/1)=2/1 Answer:(d)
Q.26
The correct statement of the following is .. [ MPPMT 1993]
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a) A body may have zero velocity but can have acceleration
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b) A body having zero velocity will necessarily have zero acceleration
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c) A boy having uniform speed can have only uniform acceleration
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d) A body having non-uniform velocity will have zero acceleration
Explanation
Answer: (a)
Q.27
If the spring extends by x on loading, then the energy stored by the spring is ( If T is the tension in the spring) ..[ [AIIMS 1997]
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a)T2 / 2x
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b) T2 / 2k
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c)2k/ T2
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d)2T2 / k
Explanation
Potential energy U=½ T x T=kx or x=T/k∴ U=½ T2 / kAnswer: (b)
Q.28
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. the area of crosscutting and the Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with time period T equal to : .. [ IIT 1993]
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a)
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b)
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c)
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d)
Explanation
The restoring force in the spring=kxFrom the formula for Young's modulus .The restoring force in wire is=YAx/L∴Therefore its spring constant is YA/ML. Now the wire and the spring both are in series. So, the equivalent spring constant is Answer: (b)
Q.29
a coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time.. [ AIIMS 2008]
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a) for amplitude of g/ω2
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b) for an amplitude of g2 / ω2
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c)at mean position of the platform
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d)at the highest position of the platform
Explanation
Fro block A to move in SHM.mg-N=mω2xwhere x is the distance from mean position. For block to leave contact N=0 ∴ mg=mω2xx=g/ω2 Answer:(a)
Q.30
The acceleration of the particle in S.H.M is .. [ MPPMT 1993]
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a) Always zero
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b) Always constant
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c) Maximum at the extreme position
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d) Maximum at the equilibrium position
Explanation
Answer: (c)
0 h : 0 m : 1 s
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