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×330⇒f=330 Hz

 



Q 2 :

The equation of a transverse wave travelling along a string is y(x,t)=4.0 sin [20×10–3x+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 (20×10–3x+600 t)

Here, ω = 600 and k = 20×10–3

∴ v=ωk=60020×10–3

          =30×10+3 mm/s=30 m/s

and 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. Which equation correctly represents this wave?          [2025]

  • y = 2 cos (0.83x – 3.35t) cm

     

  • y = 2 sin (0.83x – 3.5t) 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 (kx – ωt)

∴  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=628300m–1

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]

  • 5.0 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



Q 7 :

A steel wire with mass per unit length 7.0×10-3 kg m-1 is under tension of 70 N. The speed of transverse waves in the wire will be              [2023]

  • 200π m/s

     

  • 100 m/s

     

  • 10 m/s

     

  • 50 m/s

     

(2)

v=Tμ=7070×10-3=100 m/s



Q 8 :

A transverse harmonic wave on a string is given by y(x,t)=5sin(6t+0.003x) where x and y are in cm and t in second. The wave velocity is ______ ms-1.      [2023]



(20)

y(x,t)=5sin(6t+0.003x)

k=0.003 cm-1, ω=6 rad/s, v=ωk

⇒  60.003×102=20 ms-1



Q 9 :

The equation of wave is given by Y=10-2sin2π(160t-0.5x+π4)

where x and Y are in m and t in s. The speed of the wave is ________ km h-1.                 [2023]



(1152)

V=ωk=1600.5m/s=1600.5×185km/h=1152 km/h



Q 10 :

Two strings (A, B) having linear densities μA=2×10-4 kg/m and μB=4×10-4 kg/m and lengths LA=2.5 m and LB=1.5 m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t1 and t2 respectively to reach the joint. The ratio t1/t2 is   [2026]

  • 1.18

     

  • 1.90

     

  • 1.08

     

  • 1.67

     

(1)

Given  LA=2.5 m,  LB=1.5 m,  T=500 N

vA=TμA=5002×10-4=510×102 m/s

vB=TμB=5004×10-4=55×102 m/s

t1=LAvA=2.5510×10-2 s

t2=LBvB=1.555×10-2 s

∴  t1t2=2.5510×551.5=53×12=1.661.41=1.18



Q 11 :

The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young’s modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ________ m/s.   [2026]

  • 401

     

  • 398

     

  • 399

     

  • 402

     

(1)

Vsound=Yρ

ΔVV×100=12(ΔYY×100)-12(Δρρ×100)

=12×1-12×0.5

ΔVV×100=14

ΔV=14×V100

ΔV=1 m/s

Vfinal=400+1=401 m/s



Q 12 :

A travelling wave is described by the equation y(x,t)=0.05sin(8x-4t) m. The velocity of the wave is: - [all the quantities are in SI unit]

  • 4 ms-1

     

  • 2 ms-1

     

  • 0.5 ms-1

     

  • 8 ms-1

     

(3)

V=ωk



Q 13 :

A heavy but uniform rope of length L is suspended from a ceiling. A particle is dropped from the ceiling at the instant when the bottom end is given a transverse wave pulse. Where will the particle meet the pulse?

  • at a distance 2L3 from the bottom

     

  • at a distance L3 from the bottom

     

  • at a distance 3L4 from the bottom

     

  • at a distance L2 from the bottom

     

(2)

For dropped body L-x=12gt2, for pulse t=2xg, x is the distance from bottom free end of row.



Q 14 :

A thick uniform rope of Length L is hanging from a rigid support. A transverse wave of wavelength λ0 is set up in the middle of the rope. The wavelength of the wave as it reaches the top most point is

  • 2λ0

     

  • 2λ0

     

  • λ02

     

  • λ0

     

(2)

Here frequency f is constant. Speed of wave

v=Tμ

⇒v1v2=λ1λ2=T1T2=Mg/2Mg

⇒λ2=2λ1=2λ0



Q 15 :

The equation of a plane progressive wave is given by y=5cosπ(200t-x150) where x and y are in cm and t is in second. The velocity of the wave is _____ m/s.     [2026]

  • 120

     

  • 150

     

  • 200

     

  • 300

     

(4)

y=5cosπ(200t-x150)

v=coeff of tcoeff of x=2001150×100=300 m/s



Q 16 :

A transverse wave on a string is described by y=3sin(36t+0.018x+π4). Where x, y are in cm and t in seconds. The least distance between the two successive crests in the wave is ______ cm. (Nearest integer)   (π=3.14)                     [2026]



(349)

y=3sin(36t+0.018x+π4)

Comparing with the standard equation

y=Asin(ωt±kx+ϕ)

λ=2πk=2×3.140.018=6.280.018

λ=348.9 cm

Nearest integer=349 cm