Waves solved examples

Q1. The distance between  two consecutive crests in a wave train produced in a string is 5 cm. If 2 complete waves pass through any Point per second, the velocity of the wave is

(a) 10 cm/sec.   (b) 2.5 cm/sec

(c)  5 cm/sec.    (d) 15 cm/sec

Ans: (a) 10 cm/sec

        V=n×λ

             2×5=10

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Q2. A tuning fork makes 256 vibrations per Second in air.  When the velocity of sound is 330 m/s, then wavelength of the tone emitted is

(a) 0.56 m          (b) 0.89 m

(c)  1.11 m         (d) 1.29 m

Ans: (d) 1.29 m

        f=256Hz

        V=λ ×f

       330=256 × λ

         λ =1.28 m

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Q3. A man sets his watch by a whistle that is 2 km away.  How much will his watch be in error. ( speed of sound in air is 330 m/s)

(a) 3 second fast

(b) 3 second slow

(c) 6 second fast

(d) 6 second slow

Ans: (d) 6 second slow

        t=d/v

          =2000/330

          =6.06 ≈ 6 sec.

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Q4. When a sound wave of frequency 300 Hz passes through a medium the maximum displacement of a particle of the medium is 0.1 cm. The maximum velocity of the particle is equal to

(a) 60π cm/sec

(b) 30π cm/sec

(c) 30 cm/sec

(d) 60 cm/sec

Ans: 60π cm/sec

        Vmax = αw

                    =α × 2πn

                    = 0.1×2π×300

                    = 60π cm/sec

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Q5.  A wavelength 0.6 cm is produced in air and it travels at a speed of 300 m/s. It will be

(a) Audible wave.       (b) Infrasonic wave
(c)  Ultrasonic wave  (d) None of the above

Ans: (c) Ultrasonic wave

             n= v/λ
               = 300/0.6×10^-2 Hz
               = 3/6×10⁴  Hz
               = 50,000 Hz.   
      Hence, this wave is ultrasonic.
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Q6. The displacement y (in cm) produce by.       a simple harmonic wave is given by y =        (10/π)   sin(2000πt - πx/17). What will be.       the periodic time and maximum velocity.     of the particles in the medium ?

Ans: 
          y = (10/π) sin [2000πt - πx/17]
              Standard equation of s.h.m. is,
           y = a sin 2π [ t/T - x/λ];
       
                ∴ By comparison of the standard equation to above given equation ,
                     then we get  T = 1/1000 = 10^-3 sec.
                Particle velocity
                   = dy/dt = 10/π ✕ 2000π cos [ 2000πt - πx/17]
                 [dy/dt] = 20000 cm/s
                             = 200 m/s.
                                           
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Q7. A sound wave traveling at 350 m/s has a.           frequency of 500 Hz. What is its.                           wavelength?

Ans: 

            

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Q8. The speed of a wave in a certain medium is 900 m/s. If 3000          waves passes over a certain point of       the medium  in 2 minutes, then                 compute its wavelength?.

Ans:

     The speed of a wave in medium v = 900 ms-1

  Freq. of wave = no. of waves passing.         per sec (n) = 3000 waves/2min = 3000 /   2×60 = 25s

 Wave length = λ =?

 v = n λ

 λ  = v/n

 A = 900/25 =36m

 A = 36m

 Ans : λ = 36 m

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Q9: A wave y = a sin(ωt – kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is

(a) y = asin(ωt + kx)

(b) y = –asin(ωt + kx)

(c) y = asin(ωt – kx)

Ans

Consider option (a)

Stationary wave: Y = asin(ωt + kx) + asin(ωt – kx)

when, x = 0, Y is not zero.

The option is not acceptable.

Consider option (b)

Stationary wave: Y = asin(ωt – kx) – asin(ωt + kx) At x = 0, Y = a sinωt – asinωt = zero

This option holds good.

Option (c) gives Y = asin(ωt – kx)+ asin(ωt – kx)

Y = 2asinωt  (At x = 0, Y is not zero)

Hence only option (b) holds good

Ans: (b) y = –asin(ωt + kx)

 

Q10: The displacement y of a wave travelling in the x-direction is given by

y = 10^-4sin(600t – 2x+ π/3) metre,

where x is expressed in metre and t in second. The speed of the wave-motion, in m s^–1 is
(a) 300

(b) 600

(c) 1200

(d) 200

Ans

Given wave equation: y= 10^-4 sin (600t – 2x + π/3)m

Standard wave equation: y = asin(ωt – kx + Φ)

Compare them

Angular speed, ω = 600 sec^-1

Propagation constant, k = 2m^-1

ω/k = (2πf)/(2π/λ) = fλ = velocity

Since velocity = ω/k = 600/2 = 300 m/sec

Ans: (a) 300

 

Q11: A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

(a) 10.5 Hz

(b) 105 Hz

(c) 1.05 Hz

(d) 1050 Hz.

Ans

For the string fixed at both the ends, resonant frequency are given by f = nv/2L, where symbols have their usual meanings. lt is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these be nth and (n + 1)the harmonics.

315 = nv/2L———(1)

420 = (n+1)v/2L —(2)

Dividing equation (1) by equation (2) we get

n = 3

From equation (1), lowest resonant frequency

f₀ = v/2L = 315/3 = 105 Hz

Ans: (b) 105 Hz

Q12: A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos(αx –β t). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β in appropriate units are

(a) α = 12.50π, β = π/2.0

(b) α = 25.00π, β = π

(c) α = 0.08/π, β = 2.0/π

(d) α = 0.04/π, β =1.0/π

Ans

The wave travelling along the x-axis is given by

y(x,t) = 0.005 cos(αx – βt)

Therefore,α = k = 2π/λ As λ = 0.08 m

α = 2π/0.08 = π/0.04 ⇒ α = (π/4 ) x 100 = 25.00π

ω = β ⇒2π/2 =π

∴ α = 25.00π

β = π

Ans: (b) α = 25.00π, β = π