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Physics NEET MCQ
Oscillations And Wave Mcq
Quiz 6
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Q.1
in stationary wave .. [ BHU 1995]
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a) strain is maximum at nodes
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b) strain is maximum at anti nodes
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c)strain is minimum at nodes
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d)Amplitude is zero at all points
Explanation
Answer:(a)
Q.2
The resultant amplitude of two waves y1=a sin[ωt+(π/6)] and y2=a cosωt will be [ raj.PMT 1996]
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a) a
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b) a√2
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c) a√3
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d) 2a
Explanation
y=y1 + y2 y=a sin[ωt+(π/6)] + a cosωt y=asin&omegatcosπ/6 + a cosωtsin(π/6) + a cosωt y=(a√3 / 2) sinωt + (a/2)cosωt + a cosωt y=(a√3 / 2) sinωt + (3a/2) cosωt ∴ resultant amplitude= Answer: (c)
Q.3
two sources of nearly same frequency give beats. Which of the following statements is correct?
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a)The frequency of beats depends on the position of the observer
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b) The frequency of beats changes with time
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c)The amplitude of vibration at any point changes with frequency of the two individual sources
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d)The amplitude of vibration at any point changes with frequency equal to the difference in the frequency of the two individual sources
Explanation
Answer: (d)
Q.4
Fifty six tuning forks are arranged in a series such that each fork gives six beats per second with the previous one. Assuming the frequency of the last fork to be double of the first, the frequency of the last fork should be [ AMU 1995]
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a) 440Hz
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b) 660Hz
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c)330Hz
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d)220Hz
Explanation
Let frequency of first A=n Frequency of last tuning fork L=2n difference d=6 Number of tune forks N=56From formula L=A+(N-1)d2n=n + (56-1)6 n=330Frequency of last tuning fork=2n=2×330=660Answer: (b)
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) wave are.. [ CBSE 1994}
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a) 40 m/s
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b) 20 m/s
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c)10 m/s
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d)5 m/s
Explanation
Fixed end is a node. The next(consecutive) node is at 10cm from the fixed end. So λ/2=10cm or λ=20 cmv=n×λ v=100×20=2000 cm/secv=20 m/sec Answer:(b)
Q.6
There are three sources of sound of equal intensity and with frequencies 500, 502, 500 Hz respectively. The number of beats heard per second is .. [ ISM Dhanbad 1994]
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a) 4
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b) 2
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c) 0
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d) 3
Explanation
Let y1=asin1000πt y2=asin1004πt y3=asin1008πty=y1 + y2+y3y=asin1000πt + asin1004πt + asin1008πty=a[sin1000πt + sin1008πt + sin1004πt]y=a[sin1004πt cos4πt + sin 1004πt]y=asin1004πt[ 1 + cos4πt]∴ Amplitude 'A'=a [ 1 + cos4πt]A is maximum when cos4πt=+14πt=0, 2π, 4π, ....∴ at t=0 , 0.5, 1, 1.5 ... we hear maximum sound. The time interval is 0.5sec and the beats frequency=1/0.5=2 per secondAnswer: (b)
Q.7
Two tuning forks of frequencies 256 and 258 vibrations/second are sounded together. Then the time interval between two consecutive maxima heard by an observer is ..[ MPPMT 1988]
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a)2 second
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b) 0.5 second
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c)250 second
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d)252 second
Explanation
Number of beats per second=258-256=2 beats/sec. Thus time interval=1/2=0.5 secAnswer: (b)
Q.8
A vibrating tuning fork of frequency n is placed near the open end of long cylindrical tube. the tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75cm, the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 metre per second, then n is
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a) 500Hz
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b) 600Hz
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c)1000Hz
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d)2000Hz
Explanation
Stationary waves are formed in the cylinder. then distance between a node and antinode is λ/4=8.75 cm or λ=35.0 cm n=v / λn=350 / (35×10-2) =1000 HzAnswer: (c)
Q.9
in a resonance column, the first resonance is obtained when the level of water in the tube is at 16 cm from the open end. the next resonance point will be obtained, when the level of water from the open end is
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a) 32 cm
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b) 48 cm
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c)64 cm
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d)24 cm
Explanation
First resonance at l=λ/4=16 Second resonance at l=(3/4)λ=48 Answer:(b)
Q.10
A sound source of frequency 170 Hz is placed near a wall. A man walking from the source towards the wall finds that there is a periodic rise and fall of sound intensity. if the speed of sound in air is 340 m/s the distance separating the two adjacent positions of minimum intensity is .. [ MNR 1992]
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a) 0.5
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b) 1
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c) 3/2
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d) 2
Explanation
Stationary waves are set up in medium. the distance between adjacent nodes is λ/2 λ=v/n=340/170=2 m ∴ distance separating the two adjacent positions of minimum intensity=λ/2=1 m Answer: (b)
Q.11
In stationary waves all particles between two nodes pass through the mean position .. [ MPPMT 1999]
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a)At different times with different velocities
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b) At different time with the same velocities
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c)At the same time with equal velocity
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d)At the same time with different velocities
Explanation
Answer: (d)
Q.12
Two sound waves of relative intensities 400:1 show interference. the ratio of intensity at the maxima to minima is close to ..[ CPMT 1989]
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a) 11/9
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b) 401/399
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c)21/19
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d)√ (40/399)
Explanation
We know that intensity I ∝ a2 I1 ∝ a12I2 ∝ a22From given option 11/9 as value of 1.222Answer: (a)
Q.13
Stationary waves are formed when
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a) Two waves of equal amplitude and equal frequency travel along the same path in opposite directions
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b) Two waves of equal wavelengths and amplitude travel along the same path with equal speeds in opposite directions
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c)Two waves of equal wavelengths and equal phase travel along the same path with same speed
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d)Two waves of equal amplitude and speed travel along the same path in opposite direction
Explanation
Answer:(b)
Q.14
A wave represented by y=acos(kx-vt) is superimposed with another wave to produce stationary waves, such that x=0 is a node. The equation of the other wave is [ BHU 1998]
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a) y=a sin(kx+vt)
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b) y=-acos(kx-vt)
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c) y=-acos(kx+vt)
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d) y=-asin(kx-vt)
Explanation
to produce node at x=0 then equation must be equal to equation of wave reflected with 180° phase difference Answer: (c)
Q.15
The composition of two simple harmonic motions of equal periods at right angles to each other and with a phase difference of π results in the displacement of the particle along [ CBSE 1990]
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a)straight line
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b) circle
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c)Ellipse
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d)Figure of eight
Explanation
Answer: (a)
Q.16
The equation represents .. [ MNR 1994]
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a) transverse progressive wave
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b) longitudinal progressive wave
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c)longitudinal stationary wave
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d)transverse stationary wave
Explanation
Answer: (d)
Q.17
Stationary waves are set up in air column, velocity of sound in air is 330 m/s and frequency is 165Hz. Then distance between the nodes is .. [ EAMCET 1995]
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a) 2 m
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b) 1 m
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c)0.5 m
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d)4 m
Explanation
Wavelength of wave λ=v/n λ=330 / 165=2m Distance between two consecutive node=λ/2=2/2=1 m Answer:(b)
Q.18
Two simple harmonic motions A and B are represented by y1=a sin ( ωt + π/6)y2=a sin(ωt + 3π/6) The maximum displacement of A will be .. [APPMT 1988]
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a) At the same time as for B
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b) Earlier than that for B by T/6 second
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c)Later that for B by T/6 seconds
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d)Later than that for B by T/3 second
Explanation
For maximum displacement of A y1=a=asin(ωt1 + π/6) or ωt1 + π/6=π/2 ∴ t1=π/3ω=T/6For maximum displacement of By2=a=asin(ωt2 + 3π/6) Or ωt2+ π/2=π/2 or ωt2=0 or t2=0 ∴ Maximum displacement of A will be larger than that for B by T/6 Answer:(c)
Q.19
A stretched string is vibrating according to the equation y=5sin(πx/2)cos40πt, where y and x are in cm and t in seconds. the distance between to consecutive nodes on the string is .. [ MPPET 1999]
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a) 5 cm
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b) 4 cm
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c) 3 cm
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d) 2 cm
Explanation
Standard equation for Standing wave is y=2a sin (2πx/λ) cos(2πvt/λ) comparing given equation with standard equation we get 2π/λ=πx/2 λ=4 cm Distance between two consecutive node=λ /2=2 cm Answer: (d)
Q.20
Stationary waves are produced in medium. Which of the following statement is correct?
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a)Rarefactions occur at antinodes
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b) Condensations occur at anodes
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c)Strain at antinodes is zero
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d)Strain at antinode is maximum
Explanation
Answer: (c)
Q.21
The equation of plane progressive wave is y=0.09sin8π(t-x/20). When it is reflected at a rigid support, its amplitude becomes 2/3 rd of its previous value. The equation of the reflected wave is :
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a)
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b)
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c)
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d)
Explanation
Amplitude of the wave after reflection=0.09×(2/3)=0.06 A phase difference of π occurs due to reflection at rigid surface. Therefore the equation of reflected wave is x is +ve, as the wave is travelling along -x axisAnswer: (d)
Q.22
If two waves of the same frequency and same amplitude respectively, on superimposition produce a resultant disturbance of the same amplitude. The waves differ in phase by [ MPPMT 1990]
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a) π
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b) 2π/3
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c) π/3
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d)zero
Explanation
y1=asinωt y2=asin(ωt+Φ)y=y1+y2y=asinωt + a sub(ωt + Φ) here 2acos(Φ/2) is amplitude of resultant wave Given 2acos(Φ/2)=a⇒ φ/2=π/3φ=(2/3)π Answer:(b)
Q.23
Two vibrating tuning forks produce progressive waves given by y1=4sin(500πt) and y2=2sin(506πt) are held near the ear of a person. if the number of beats heard per second be b and the ratio of maximum to minimum intensity be A, then [ CPMT 1988]
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a) B=3 and A=2
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b) B=3 and A=9
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c) B=6 and A=2
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d) B=6 and A=9
Explanation
Number of beats=(506-500)/2=3 Imax=(a1 + a2)2 Imax=(4+2)2=36 Imin=(4-2)2=4 Thus Imax / Imin=36/4=9 Answer: (b)
Q.24
The equation of wave traveling in a string can be be written as y=3cosπ(100t-x). Its wavelength is ..[ MNR 1985]
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a)100 cm
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b) 2 cm
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c)5 cm
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d)None of the above
Explanation
Comparing given wave equation with we get 2π/λ=π or λ=2cm Answer: (b)
Q.25
Transverse waves can propagate .. [ CPMT 1984]
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a) Both in a gas and a metal
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b) In a gas but not in a metal
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c)Not in a gas but in metal
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d)Neither in a gas nor in metal
Explanation
Answer: (c)
Q.26
The displacement y ( in cm) produced by a simple harmonic wave is given by y=(10/π) sin(2000πt - πx/17) The periodic time and maximum velocity of the particles in the medium will respectively be ... [ CPMT 1986]
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a)10-3 sec and 340 m/s
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b) 10-4 and 20 m/s
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c)10-3 and 200 m/s
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d)10-2 and 2000,/sec
Explanation
Comparing given equation with standard equation y=asin(ωt - kx) we have 2000π=ω ∴ T=2π / ω=2π/2000π=10-3 sec Vmax=2×102 m/sec Answer:(c)
Q.27
The velocities of sound of an ideal gas at temperature T1 and T2 K are found to be V1 and V2 respectively. If the root mean square speeds of the same gas at the same temperature T1 and T2 are v1 and v2 respectively then .. [ CPMT 1986]
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a) v2 / v1=V2 / V1
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b) v2 / v1=V1 / V2
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c) v2 / v1=[ (√V2) / V1]
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d) v2 / v1=[ (√V1) / V2]
Explanation
Ratio of velocity of sound at two temperatures is v1 / v2=√( T1/T2) Ratio's of rm.s velocity at two temperatures is V1 / V2=√( T1/T2) ∴ v1 / v2=V1 / V2 Answer: (a)
Q.28
A transverse wave is described by the equation Y=Yosin2π(ft- x/λ). The maximum particle velocity is equal to four times the wave velocity if ... [ MPPMT 1998]
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a)λ=πYo/4
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b) λ=πYo/2
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c)λ=πYo
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d)λ=2πYo
Explanation
Y=Yosin2π(ft- x/λ) Wave velocity=fλMax. particle velocity=aω=Yo2πf=2πYof 2πYof=4fλ Or λ=πYo / 2 Answer: (b)
Q.29
Which of the following properties of sound is affected by change in air temperature ..[ AMU 1995]
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a) Amplitude
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b) Frequency
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c)Wavelength
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d)Intensity
Explanation
Answer: (c)
Q.30
Ultrasonic waves are used for stirring liquid solution because
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a) They don not produce noise during the operation
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b) They are easy to produce
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c)They can produce perfectly homogeneous solutions
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d)They do not produce chemical reactions in the solutions
Explanation
Answer:(c)
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