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Q1. If the speed of the wave shown in the figure is 330m/s in the given medium, then the equation of the wave propagating in the positive x-direction will be (all quantities are in M.K.S. units)
•  y=-0.05 sin⁡〖2π(3300 t+10 x)〗
• y=0.05 sin⁡〖2π(3300 t+10 x)〗
•  y=-0.05 sin⁡〖2π(3300 t-10 x)〗
•  y=0.05 sin⁡〖2π(3300 t-10 x)〗
Solution
Here A=0.05m,5λ/2=0.25⇒λ=0.1m Now standard equation of wave y=A sin⁡〖2π/λ〗 (vt-x) ⇒y=0.05 sin⁡〖2π(3300t-10x)〗

Q2.What is the phase difference between two successive crests in the wave?
•
•
•  π/4
•  π
Solution
Relation of path difference and phase difference is given by ∆Φ=2π/γ×∆x Where ∆x is path difference. But path difference between two crests ∆x=λ Hence, ∆Φ=2π/λ×λ=2π

Q3.  Two identical plain wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously
•   0.04
•  0.03
•  0.01
•  0.02
Solution

Q4. Two strings X and Y of a sitar produce a beat frequency 4Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was
•  304 Hz
•  302 Hz
•  298 Hz
•  296 Hz
Solution

Q5.Two waves of wavelength 1.00m and 1.01m produces 10 beats in 3s. What is the velocity of the wave?
•  336.6 ms^(-1)
•  200 ms^(-1)
•  115.2 ms^(-1)
•  150 ms^(-1)
Q6. Two waves are approaching each other with a velocity of 16 m/s and frequency n. The distance between two consecutive nodes is
•  16/n
•  8/n
• n/8
•  n/16
Solution
Distance between two nodes =λ/2=v/2n=16/2n=8/n

Q7.When two sound waves are superimposed, beats are produced when they have
•  Different amplitudes and phase
•  Different velocities
•  Different frequencies
•  Different phases
Solution
For producing beats, their must be small difference in frequency

Q8.It takes 2.0 s for a sound wave to travel between two fixed points when the day temperature is 10°C. if the temperature rises to 30°C the sound wave travels between the same fixed parts in
•  2.0s
•  1.9s
•  2.1s
•  2.2s
Solution
Q9.From a point source, if amplitude of waves at a distance r is A, its amplitude at a distance 2r will be
•  2A
•  A
•  A/4
•  A/2
Solution
Intensity=energy/sec/area=power/area. From a point source, energy spreads over the surface of a sphere of radiusr. Intensity =P/A=P/(4πr^2 )∝1/r^2 But Intensity=(Amplitude)^2 ∴(Amplitude)^2∝1/r^2 or Amplitude∝1/r At distance 2r,amplitude becomes A/2

Q10. A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats per second. The beat frequency does not change if the length of the wire is changed to 21 cm. The frequency of the tuning fork (in Hertz) must be
•  220
•  205
•  210
• 215
Solution
Let the frequency of tuning fork be N As the frequency of vibration string ∝1/(length ofstring) For sonometer wire of length 20 cm, frequency must be (N+5) and that for the sonometer wire of length 21cm, the frequency must be (N-5) as in each case the tunning fork produces 5 beats/sec with sonometer wire Hence n_1 l_1=n_2 l_2⇒(N+5)×20=(N-5)×21 ⇒N=205 Hz ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: WAVE MOTION QUIZ-10
WAVE MOTION QUIZ-10
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