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Oscillations And Wave Mcq
Quiz 1
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Q.1
In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic? [ CBSE-PMT 1996]
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a) 0
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b) 1/4
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c)1/2
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d)3/4
Explanation
Total energy of particle executing S.H.M of amplitude (A)E=½ ( m ω2A2)Kinetic energy of particle when x=A/2 is Clearly, KE/ total energy=3/4 Answer:(d)
Q.2
A cylindrical resonance tube, open at both ends, has a fundamental frequency f in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be...[ AFMC 1997]
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a)3f/2
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b) 2f
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c)f
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d)f/2
Explanation
from figure it is clear that in both the cases wavelength is same. So frequency of fundamental tone will also be same. Hence answer is (c)Answer: (c)
Q.3
A wave of frequency 500 Hz has a velocity of 350 m/s. The distance between two nearest points, if the wave is 60° out of phase, will be approximately...[ AFMC 1997]
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a) 70 cm
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b) 0.7 cm
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c)12.0 cm
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d)120. cm
Explanation
60°=π/3 if phase difference is 2Π=λ path differenceif phase difference is π/3=?path difference=(π/3×λ) / 2Π path difference=λ / 6 --eq(1)Wave length of wave=velocity / frequency wave length=350/500=0.7 m=70 cm substituting value of wave length in equation 1, we get path difference=70/6=11.66 cm Answer: (c)
Q.4
Energy of simple harmonic motion depends upon. [ AFMC 1997]
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a) 1/ω2
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b) ω
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c)a2
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d)1/ a2
Explanation
Energy of simple harmonic motion is proportional to square of its amplitude Answer:(c)
Q.5
A wave of frequency 100Hz is sent along a string towards a fixed end. When this wave travels back, after reflection, a node is formed at a distance of 10cm from the fixed end of the string. The speeds of incident ( and reflected ) waves are [ AFMC 1997]
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a) 48 m/s
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b) 20 m/s
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c) 10/m/s
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d) 15 m/s
Explanation
Node is formed at 10 cm hence λ/2=10 λ=20cm=0.2 m Velocity of wave v=λ×frequency v=0.2×100=20 m/sec Answer: (b)
Q.6
A source of sound is traveling with a velocity 40km/he towards an observer and emits sound of frequency 2000 Hz. If velocity of sound is 1220 km/hr, then the apparent frequency heard by an observer is ...[ AFMC 1997]
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a) 1980 Hz
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b) 1950 Hz
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c) 2068 Hz
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d) 2080 Hz
Explanation
From the formula here f' is apparent frequency, f is the frequency of source= 2000Hz, 'v' is velocity of sound =1220m/s, vs velocity of source = 40 m/s substituting values in above equation we get Answer: (c)
Q.7
A glass rod 20cm long is clamped at the middle. It is set into longitudinal vibration. If the emitted sound frequency is 4000Hz, the velocity of sound in glass will be..[ AFMC 1997]
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a) 2800 m/s
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b) 3200 m/s
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c) 1600 m/s
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d) 2000 m/s
Explanation
At the point of clamping node will be produce and half wave length will be equal to length of rod as shown in figure L = λ/2 0.2=λ/2 λ =0.4 m Velocity of sound = frequency ×Wave length Velocity of sound = 4000×0.4= 1600 m/s Answer: (c)
Q.8
Which of the following does not show polarization? [ AFMC 1997]
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a) Transverse wave in gas
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b) longitudinal in gas
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c) both (a) and (b)
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d) none of these
Explanation
Longitudinal wave do not show polarization because vibration of gas molecules is along the direction of propagation Answer:(b)
Q.9
A point mass m is suspended at the end of a mass-less wire of length L and cross-sectional area A. If Y is the Young's modulus of the wire, then the frequency of the oscillation for simple harmonic oscillation along the vertical direction is ....[ AFMC 1998]
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a)
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b)
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c)
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d)
Explanation
According to equation Here force F is restoring force of the wire acts on mass'm' F = Y(A) (Δl/l) If 'a' is acceleration then ma = Y(A) (Δl/L) So acceleration ∝ displacement Δl The motion of mass will perform SHM Answer: (c)
Q.10
A body is executing simple harmonic motion with an angular frequency of 2 rad/sec. The velocity of the body at 20mm displacement, when the amplitude of the motion is 60mm is [ AFMC 1998]
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a)131 mm/s
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b) 188 mm/s
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c)113 mm/s
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d)90 mm/s
Explanation
Formula for velocityω=2 rad/sec, a=60 mm , x=20 mmby substituting the values in above equation we getv=113 mm/secAnswer: (c)
Q.11
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β, ten its time period of vibration will be...[ AFMC 1998]
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a) (2πβ) / α
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b) β2 / α2
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c)α/ β
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d)β2 / α
Explanation
Maximum acceleration α=ω2 aMaximum velocity β=ωaα / β=ω2 a / ωaα / β=ωα / β=2π / T T=2π × β / α Answer: (a)
Q.12
A hole is bored along the diameter of earth and a stone is dropped into the frictionless tunnel. If the radius of earth is R, then the time period of executing simple harmonic motion is ...[ AFMC 1998]
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a)
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b)
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c)
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d)None of these
Explanation
If we drop a stone into the tunnel, it will be attracted towards the center. When it reaches the center, then the force acting on it becomes zero, due to inertia of motion it ill cross the center and go to the other side of the center, again the center starts pulling it towards itself. In this way a force acts on it towards a fixed point. In this way it will execute SHM. We know that acceleration due to gravity inside the earth is given byhere x is displacementF=mg' then mg'=mg(x/ R)g'=(g/R) x acceleration ∝ displacement (x) so ω2=g/R Answer:(c)
Q.13
The driver of a car traveling with speed 30 m/s towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by the driver is ...[ AFMC 1998]
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a) 500Hz
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b) 550Hz
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c) 720 Hz
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d) 555 Hz
Explanation
Here source and observer are virtually approaching to each other, so here f' is apparent frequency, f is actual frequency of sound.v is the velocity of sound in air, u is the velocity of observer Answer: (c)
Q.14
A particle moving along the x-axis, executes simple harmonic motion then the force acting on it is given by [ CBSE-PMT 1988]
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a)-A kx
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b) A cos(kx)
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c)A exp(-kx)
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d)A kx
Explanation
for simple harmonic motion, F=-kxHere k=AkAnswer: (a)
Q.15
If a simple harmonic oscillation has got a displacement of 0.02 m and acceleration equal to 2.0m/s2 at any time, the angular frequency of the oscillator is equal to .. [ CBSE-PMT 1992]
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a) 10 rad/s
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b) 0.1 rad/s
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c)100 rad/s
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d)1 rad/s
Explanation
acceleration=displacement × ω2 ∴ ω2=acceleration / displacement ω2=2.0 / 0.02 ω 2=100 ω=10 rad/sAnswer: (a)
Q.16
A body executes S.H.M 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 .. [ CBSE-PMT 1993]
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a) A/4
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b) A/2
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c) 3A/4
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d) Some other fraction of A
Explanation
P.E=½ mω2x2 Total energy E=½ m ω2A2 P.E=¼ E ⇒ x=½ A Note The displacement at which the P.E becomes n times the total energy is given by x=A √n Answer: (b)
Q.17
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 .. [ CBSE-PMT 2010]
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a)T
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b) T / √2
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c)2T
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d)T√2
Explanation
Since spring is same thus T ∝ m1/2Thus T'=T√2Answer: (d)
Q.18
A particle executing simple harmonic motion of amplitude 5cm has maximum speed of 31.4cm/s. The frequency of its oscillation is ... [ CBSE-PMT 2005]
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a) 4Hz
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b) 3Hz
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c)2Hz
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d)1Hz
Explanation
In S.H.M vmax=A ω vmax=A( 2π f)f=vmax / 2πAf=31.4 / 2(3.14)×5=1HzAnswer: (d)
Q.19
Two simple harmonic motions od angular frequency 100 and 1000 rad s-1 have the same displacement amplitude. The ratio of their maximum acceleration is . [ CBSE-PMT 2008]
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a)1: 10
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b) 1: 102
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c)1:103
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d)1:104
Explanation
acceleration a ∝ ω2 on taking the ratio of acceleration we get 1:102 Answer:(b)
Q.20
Two springs of spring constant k1 and k2 are joined in series. The effective spring constant of the combination is given by . [ CBSE-PMT 2004]
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a) k1k2 / (k1 +k2)
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b) k1k2
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c) (k1+k2) / 2
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d) k1 + k2
Explanation
Answer: (a)
Q.21
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 .. [ CBSE-PMT 2007]
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a)T/8
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b) T/12
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c)T/2
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d)T/4
Explanation
Displacement from the mean position According to problem y=a/2This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium positionAnswer: (b)
Q.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 path of the two particles. The phase difference is .. [ CBSE-PMT 2011]
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a) 0
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b) 2π / 3
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c)π
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d)π / 6
Explanation
Equation for SHM is given by y=A sin(ωt + δ), Here (ωt + δ) is called phase When y=A/2, then sin(ωt + δ)=1/2Thus (ωt + δ)=π /6Now for 1st Phase (ωt + δ)=Φ1=Φ / 6For second particle when y=A/2Φ2=Φ - π / 6=5π / 6 ∴ Phase difference Φ=Φ2 - Φ1=2π / 3Answer: (b)
Q.23
Which of the following equations of motion represents simple harmonic motion? [ CBSE-PMT 2009]k, k0, k1 and 'a' are all positive
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a)Acceleration=-k(x+a)
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b) Acceleration=k(x+a)
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c)Acceleration=kx
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d)Acceleration=-k0x + k1x2
Explanation
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.. Answer:(a)
Q.24
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.. [ CBSE-PMT 2003]
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a) 2T
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b) T/4
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c) 2
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d) T/2
Explanation
Since mass attached to spring is constant thus Periodic Time T ∝ 1 / √k When spring is cut in four parts new spring constant k=4k Thus new Periodic time will be T/2 Answer: (d)
Q.25
A block of mass M is attached to the lower end of a vertical spring. The spring is hung from ceiling and has force constant value 'k'. The mass is released from the rest when the spring initially un stretched. The maximum extension produced in the length of the spring will be.. [ CBSE-PMT 2009]
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a)2Mg/k
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b) 4Mg/k
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c)Mg/2k
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d)Mg/k
Explanation
If 'x' is the extension produced in the springforce acting upward is kx , while force acting downward is Mg At equilibrium kx=Mgx=Mg/kAnswer: (d)
Q.26
Which of the following is a simple harmonic motion? [ CBSE-PMT 1994]
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a) Ball bouncing between two rigid vertical walls
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b) Particle moving in a circle with uniform speed
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c)Wave moving through a string fixed at both ends
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d)Earth sinning about its own axis
Explanation
A wave moving through a string fixed at both ends, is a transverse wave formed as result of simple harmonic motion of particles of the string..Answer: (c)
Q.27
If length of a simple pendulum is increased by 2%, then the time period .. [ CBSE-PMT 1997]
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a)increases by 2%
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b) decreases by 2%
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c)increases by 1%
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d)decreases by 1%
Explanation
We know that periodic time T ∝ √ l∴ If length is increased by 2%, time period increases by 1% Answer:(c)
Q.28
Which of the following statement is true for the speed v and the acceleration a of a particle executing simple harmonic motion? [ CBSE-PMT 2004]
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a) When v is maximum, a is zero
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b) When v is maximum,a is maximum
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c) Value of a is zero, whatever may be the value of v
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d) When v is zero, a is zero
Explanation
At equilibrium position velocity is maximum but acceleration is zero. Answer: (a)
Q.29
Two simple harmonic motions with the same frequency act on a particle at right angles i.e. along x axis and y axis. If the two amplitudes are equal and the phase difference is π/2, the resultant motion will be.. [ CBSE-PMT 1997]
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a)a circle
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b) an ellipse with the major axis along y-axis
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c)an ellipse with the major axis along x-axis
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d)a straight line inclined at 45° to the x-axis
Explanation
Equation of two simple harmonic motions y=Asin(ωt + Φ) --(1) x=Asin(ω t + Φ + π /2) x=Acos (ωt + Φ) --(2)on squaring and adding equation (1) and (2) we getx2 + y2=A2This is an equation of circle. Hence, resulting motion will be a circular motion.Answer: (a)
Q.30
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 [ CBSE-PMT 2008]
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a) T/8
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b) T/6
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c)T/3
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d)T/12
Explanation
According to questioncomparing above equation we get cos ( ωt + π/6)=1/2⇒ ωt + π/6=π/3 ∴ t=T / 12Answer: (d)
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