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)
Solution
Here A=0.05m,5λ/2=0.25⇒λ=0.1m Now standard equation of wave y=A sin〖2Ï€/λ〗 (vt-x) ⇒y=0.05 sin〖2Ï€(3300t-10x)〗
Here A=0.05m,5λ/2=0.25⇒λ=0.1m Now standard equation of wave y=A sin〖2Ï€/λ〗 (vt-x) ⇒y=0.05 sin〖2Ï€(3300t-10x)〗
Q2.What is the phase difference between two successive crests in the wave?
Solution
Relation of path difference and phase difference is given by ∆Φ=2Ï€/γ×∆x Where ∆x is path difference. But path difference between two crests ∆x=λ Hence, ∆Φ=2Ï€/λ×λ=2Ï€
Relation of path difference and phase difference is given by ∆Φ=2Ï€/γ×∆x Where ∆x is path difference. But path difference between two crests ∆x=λ Hence, ∆Φ=2Ï€/λ×λ=2Ï€
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
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
Solution
Q5.Two waves of wavelength 1.00m and 1.01m produces 10 beats in 3s. What is the velocity of the wave?
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
Solution
Distance between two nodes =λ/2=v/2n=16/2n=8/n
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
Solution
For producing beats, their must be small difference in frequency
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
Solution
Q9.From a point source, if amplitude of waves at a distance r is A, its amplitude at a distance 2r will be
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/(4Ï€r^2 )∝1/r^2 But Intensity=(Amplitude)^2 ∴(Amplitude)^2∝1/r^2 or Amplitude∝1/r At distance 2r,amplitude becomes A/2
Intensity=energy/sec/area=power/area. From a point source, energy spreads over the surface of a sphere of radiusr. Intensity =P/A=P/(4Ï€r^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
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
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
