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This is one of our many quizzes on Wave properties. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Wave properties Quiz 2

physics

 A cargo ship reports rogue waves headed to the shore which is $156\,m$ away from ship at a velocity of $6.5\,ms^{-1}$. If the ship sends a SOS signal to a lighthouse on the shore, how long does the lighthouse have to alert the people to evacuate the beach? $\text{(Take velocity of ‘O’ signal to be }340\,ms^{-1})$ 

First find the time in which the rogue waves will reach the shore and then account for the time taken by the SOS signal to reach the shore to arrive at the evacuation time.  

  We are given that for the rogue wave: $ \\ v = 6.5\,ms^{-1}$, and $d = 156\,m \\ $ The time in which the rogue waves will reach the shores will be: $ \\ t = \dfrac{156}{6.5} \\ \Rightarrow t = 24\,s \\ $ Now, given that the velocity of the SOS signal is $v^{\prime} = 340\,ms^{-1}$ and travels a distance $d = 156\,m$, the time it takes for the SOS signal to reach the lighthouse will be: $ \\ t^{\prime} = \dfrac{d}{v^{\prime}} \\ \Rightarrow t^{\prime} = \dfrac{156}{340} \\ \Rightarrow t^{\prime} = 0.46\,s \\ $ Therefore, the lighthouse will have a time of $T$ as, $ \\ T = t-t^{\prime} \\ \Rightarrow T = 24-0.46 \\ \therefore T = 23.54\,s \approx 23.5\,s $

The progression of an ocean wave is as shown in the diagram below. Determine the velocity of the wave. If a surfer wants to get from point $B$ to point $P$, with what minimum velocity must he paddle before he can ride the wave at point $P$? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1289567d-5b52-48ff-8ace-1e0f8a4e8db48530264519933635579.png"/> 

Find the frequency of the wave and its wavelength from the figure to find its velocity. Calculate the surfer’s velocity by considering the distance over the wave that he needs to travel.  

  The progression of the ocean wave is described by the equation: $ \\ y = A_0sin(4\pi t) \\ $ The angular frequency is $ \\ \omega = 4\pi \\ $ If $f$ is the frequency of vibration then: $ \\ 2\pi f = 4\pi \\ \Rightarrow f = 2Hz \\ $ From the figure, we see that the distance between the first two consecutive crests, equivalent to the wavelength is: $ \\ \lambda = 1.0-0.3 \\ \Rightarrow \lambda  = 0.7\,m \\ $ Therefore, the velocity of the wave is: $ \\ v = f\lambda \\ \Rightarrow v = 2 \times 0.7 \\ \Rightarrow v = 1.4\,ms^{-1} \\ $ Now, for the surfer to reach point P from B he must paddle a distance of $d = 1.7 – 0.3 = 1.4\,m$ $ \\ $ This means that he must paddle at a velocity of: $ \\ v = f \times d \\ \Rightarrow v = 2 \times 1.4 \\ \therefore v = 2.8\,ms^{-1}$

 You decide to go on a fishing tour to Dal Lake in Srinagar. While you are seated in the middle of the lake in your boat, waiting endlessly for the fishing line to wobble under the bait, you start to count the number of waves that pass the fishing line at the surface, for every ten seconds, four crests pass the fishing line, with each wave having a length of not more than $1.5\,m$. How fast were the waves moving? 

Determine the frequency of the wave from the given number of crests in a number of seconds to arrive at the appropriate speed of the progressive wave.  

 Given that four wave crests pass the line in ten seconds, the frequency of the wave will be: $ \\ f = \dfrac{4}{10} \\ \Rightarrow f  = 0.4Hz \\ $ <br>The wavelength of the passing wave is $1.5 m$ $ \\ $ In such a case, the speed of the passing wave will be: $ \\ v = f\lambda \\ \Rightarrow v = 0.4 \times 1.5 \\ \therefore v = 0.6\,ms^{-1}$

You and your friend find an old abandoned underground railroad in your town. While exploring the track in the dark you get separated from your friend. Your friend decides to strike the railroad track so you can hear the propagated wave through the track. You put your ear against the railroad track and as soon as you hear it you strike the track again, and your friend records that it takes $0.8\,s$ for the sound to travel back and forth. If the speed of the propagated wave in steel is $5000\,ms^{-1}$, how far are you from your friend? 

The time that your friend records will be the time taken by the wave to travel twice the distance between you and your friend.  

 Let the distance between you and your friend be $d$.  $ \\ $ The sound takes a time of $0.8\,s$ to travel a distance of $d^{\prime} = 2d$. $ \\ $ Given that $v = 5000\,ms^{-1}$ $ \\ d^{\prime} = v \times t \\ \Rightarrow 2d = 5000 \times 0.8 \\ \therefore d = 2000m = 2\,km$

 The seismic waves that travel through the interior of the Earth are of two types: P-type $\text{(primary or pressure)}$ waves are capable of propagating through gases, liquids and solids and are longitudinal waves whereas S-type $\text{(secondary or shear)}$ waves propagate only through solids or high viscosity liquids and are transverse waves. Suppose that the average density of the interior of the earth is around $\rho = 2.4\,gcm^{-3}$, and its bulk modulus is $B = 24\,GPa$. If the modulus of rigidity of the earth’s interior is $\mu = 21\,GPa$, determine the velocity with which the P and S- waves would propagate through the interior of the earth.  

The P-waves travel through the bulk of the earth producing compressive changes whereas the S-waves are propagated perpendicular to the bulk of the earth producing torsional changes. 

 Given that $\rho = 2.4\,gcm^{-3} = 2.4 \times 10^3\,kgm^{-3}$, $B = 24\,GPa = 24 \times 10^9\,Pa$ and, $ \Rightarrow\mu = 21\,GPa = 21 \times 10^{9}\,Pa $ $ \\ $ The velocity of the P-waves is given by the velocity of compressions of any longitudinal wave through a bulk of solid: $ \\ v = \sqrt{\dfrac{B+\left(\dfrac{4}{3}\right)\mu}{\rho}} \\ \Rightarrow v = \sqrt{\dfrac{24 \times 10^9+\left(\dfrac{4}{3}\right) \times 21 \times 10^9}{2.4 \times 10^{3}}} \\ \Rightarrow v = \sqrt{\dfrac{(24+28) \times 10^{9}}{2.4 \times 10^3}} \\ \Rightarrow v = 4.65 \times 10^3\,ms^{-1} \\ \Rightarrow v = 4.65\,kms^{-1} \\ $ The velocity of S-waves is given by the velocity of crests of any transverse wave through the solid surface: $ \\ v = \sqrt{\dfrac{\mu}{\rho}} \\ \Rightarrow v = \sqrt{\dfrac{21 \times 10^9}{2.4 \times 10^3}} \\ \Rightarrow v = 2.95 \times 10^3\,ms^{-1} \\ \therefore v = 2.95\,kms^{-1}$