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      Physics (Wave)

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

      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

      Wavelength = λ =?

      v = n λ

      λ = v/n

      A = 900/25 =36m

      A = 36m

      Answer : λ = 36 m

      2. Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.


      Helium – 2 mole, oxygen – 4 mol

      He & O2 are mixed, hence the molecular weight of the mixture of gases

      Answer: 400.9 ms-1

      3. A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bedrock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between the two cases.


      The velocity of SONAR waves in water C = 1500 ms-1

      Time is taken by wave after reflection from the bottom of the sea

      2t = 3.5s

      t =1.75s

      Distance covered (d) =?

      C = d/t => d = c.t = 1500 × 1.75 – 2625 m

      After moving 100km

      The time taken by the wave = 2t = 2s

      T = 2/2 = 1s

      d =?

      d = 1500 × l

      = 1500

      The difference between these two heights. = 2625 – 1500

      = 1125m

      Answer : Δd = 1149.75 m

      4. A sound wave is transmitted into a tube as shown in the figure. The sound wave splits into two waves at point A which recombine at point B. Let R be the radius of the semi-circle which is varied until the first minimum. Calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m.

      Answer : R = 21.9 m


      Consider the sound waves starting from pt A. The sound waves will meet in picture B & interfere and can be detected.

      The path length of sound waves passing through the curve is equal to half the circle having radio centred at c.

      Thus, be path length of be sound waves in the curve part is

      L1 = πγ

      The path length of the sound in tube ABC is equal to the diameter of the circle having radius γ centred at e.

      So the path length of the sound waves travelling in it = L2 =2r

      The path difference of the sound waves in the picture is

      ΔP = L1 – L2

      = πr— 2r

      ΔP = r(π – 2) …(1)

      For the sound to be heard minimum at the detected the difference in path length of be sound waves is

      ΔP = λ/2 …………(2)

      From equation (1)

      r(π- 2) = λ/2

      wavelength of sound = λ = 50m

      the radius of being semicircle r =?

      r = 50 / [2 × 3.14 – 4] = 50/2.28

      r = 21.9m

      Answer : R = 21.9 m

      5. N tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. If the last fork gives double the frequency of the first (called an octave), Show that the frequency of the first tuning fork is f = (N−1)n.


      No.of tuning forks = N

      The frequency of the first is double to be last

      The difference of frequency for two successive tuning forks NHz

      Let be the frequency of the first tuning fork be f

      Then be the frequency of the last tuning fork is f(N-l)n

      frequency is reduced to are fourth

      flast = – f /2

      an = a + (n-1)d

      (i.e.) 2f = f + (N-1)n

      = (N-1)n

      f = (N-1)n

      6. Let the source propagate a sound wave whose intensity at a point (initially) is I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point.


      The intensity of sound wave = Iold

      Amplitude is doubled = 2A = Anew

      Frequency is produced to our fourth

      Answer: Inew ∝ ¼ I old


      7. Consider two organ pipes of the same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of the closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.

      Answer: 500Hz


      8. A police in a siren car moving with a velocity of 20 ms-1 chases a thief who is moving in a car with a velocity of v0ms-1. The police car sounds at a frequency 300Hz, and both of them move towards a stationary siren of a frequency 400Hz. Calculate the speed at which the thief is moving.


      The velocity of the car = 20ms-1

      The frequency of the car = 300Hz

      The frequency stationary siren = 400Hz

      The speed at which the thief moving =?

      Answer: vthief = 10 m s-1

      9. Consider the following function

      (a) y = x2 + 2 α t x

      (b) y = (x + vt)2

      which among the above function can be characterized as a wave ?.


      (a) y = x2 + 2 a t x

      It is not describing a wave.

      (b) y = (x + 2 vx)2

      It satisfies the wave equation.

      Answer: (a) function is not describing wave (b) satisfies wave equation.


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