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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)
Solution
 


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

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