Q 1 :    

A plane progressive wave is given by y=2 cos 2π(330t-x)m. The frequency of the wave is                      [2024]

  • 340 Hz

     

  • 165 Hz

     

  • 330 Hz

     

  • 660 Hz

     

(3) 

        y=2 cos 2π(330t-x)m

         y=A cos (ωt-kx)

         Comparing ω=2π×330

          2πf=2π×330f=330 Hz

 



Q 2 :    

The equation of a transverse wave travelling along a string is y(x,t)=4.0 sin [20×103x+600 t] mm, where x is in the mm and t is in second. The velocity of the wave is           [2025]

  • + 30 m/s

     

  • – 60 m/s

     

  • – 30 m/s

     

  • + 60 m/s

     

(3)

y=4 sin [(20×103x+600 t)

Here, ω = 600 and k20×103

 v=ωk=60020×103

          =30×10+3 mm/s=30 m/s

abd direction is towards –ve x-axis

  v = – 30 m/s



Q 3 :    

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x = 0 at t = 0 sec and maximum displacement of the wave is 2 cm. With equation correctly represents this wave?          [2025]

  • y = 2 cos (0.83x – 3.35t) cm

     

  • y = 2 sin (0.83x – 3.35t) cm

     

  • y = 2 cos (3.35x – 0.83t) cm

     

  • y = 2 cos (0.13x – 0.5t) cm

     

(1)

v=distancetime

v=1.20.3=4 cm/s

k=2πλ=2π7.5=4π15=0.83

v=ωk  ω=vk=4×4π15=3.35

At t = 0, x = 0, there is a crest

So, y = A cos (kxωt)

  y = 2 cos (0.83x – 3.35t) cm



Q 4 :    

Displacement of a wave is expressed as x(t)=5 cos (628t+π2) m. The wavelength of the wave when its velocity is 300 m/s is          [2025]

  • 5 m

     

  • 3 m

     

  • 0.5 m

     

  • 0.33 m

     

(2)

x(t)=5 cos [628t+π2] m

Wave velocity v = 300 m/s

v=ωk

300=628K  K=628300m1

2πλ=628300  λ=2×3.14×300628

 λ=3 m



Q 5 :    

The equation of a wave travelling on a string is y = sin [20 πx + 10 πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :          [2025]

  • 0.5 cm

     

  • 20 cm

     

  • 10 cm

     

  • 2.5 cm

     

(1)

k=20π=2πλ  λ=10 cm

Minimum distance between 2 points having same speed is λ2.

Minimum distance =λ2=5 cm



Q 6 :    

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is v1, and that in the other string having radius of cross section R/2 is v2. Then v2v1=          [2025]

  • 2

     

  • 2

     

  • 8

     

  • 4

     

(2)

v=Tμ=TρπR2

 v2v1=R1R2=2