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

Wave equation Quiz 2

physics

 Mehul found that two waves superposed to give a resultant as$y=10\cos 50x\cos 4t$. He assumed that one of the waves was having an equation given as, ${{y}_{1}}=5\sin \left( 50x+4t \right)$.Find out the other wave equation according to Mehul’s findings. 

 We may add two waves that are in the same phase together to get the resultant of the wave as they superpose. 

  $\text{According to the superposition }\!\!'\!\!\text{ s principle,} \\ y={{y}_{1}}+{{y}_{2}}.............(1) \\ \text{It has been given that,} \\ {{y}_{1}}=5\sin \left( 50x+4t \right) \\ \text{Substituting the values in equation (1),} \\ 10\sin 50x\cos 4t=5\sin \left( 50x+4t \right)+{{y}_{2}} \\ \Rightarrow {{y}_{2}}=5\sin \left( 50x+4t \right)-10\sin 50x\cos 4t \\ \text{Using the trigonometric identity,} \\ \sin \left( A+B \right)+\sin \left( A-B \right)=2\sin A\cos B \\ \text{Comparing the equations,} \\ A=50x \\ \Rightarrow B=4t \\ \text{Substituting,} \\ \Rightarrow {{y}_{2}}=5\sin \left( A-B \right) \\ \therefore {{y}_{2}}=5\sin \left( 50x-4t \right)$

 A particle producing sound waves which was moving through a medium of gas of density $400\text{ }kg{{m}^{-3}}$ with a velocity mentioned as $331\text{ }m{{s}^{-1}}$. Calculate the number of moles of the gas medium at a temperature $30\text{ }K$ and mass $4\text{ }g$. Let us assume that the Laplace correction is $1$. 

 The velocity of sound can be defined as the square root of the ratio of the product of the Laplace correction and the pressure to the density. 

  &nbsp;$\text{The velocity of sound,} \\ \text{v=}\sqrt{\dfrac{\gamma P}{\rho }} \\ \text{Rearranging the terms,} \\ P=\dfrac{{{v}^{2}}\rho }{\gamma } \\ \text{Substituting the values,} \\ P=\dfrac{{{\left( 331 \right)}^{2}}\times 400}{1} \\ \Rightarrow P=4.4\times {{10}^{7}}\text{ }atm \\ \text{According to the ideal gas equation,} \\ PV=NRT \\ \text{Rearranging the equation,} \\ N=\dfrac{PV}{RT}...............(1) \\ \text{The density can be written as,} \\ \rho =\dfrac{M}{V} \\ \Rightarrow V=\dfrac{M}{\rho } \\ \text{Substituting the values,} \\ V=\dfrac{4\times {{10}^{-3}}}{400} \\ \Rightarrow V={{10}^{-5}}\text{ }{{m}^{-3}} \\ \text{Substituting this in equation (1),} \\ N=\dfrac{4.4\times {{10}^{7}}\times {{10}^{-5}}}{8.314\times 30} \\ \therefore N=1.76$

 A body was applied with a stress mentioned as $300\text{ }N{{m}^{-2}}$ and the value of the strain acting on the body can be written as $60$. Calculate the density of the body if the velocity of the sound produced by the body is $331\text{ }m{{s}^{-1}}$. 

 The velocity of the sound can be found by taking the square root of the ratio of the young’s modulus to the density of the body. 

 $\text{The equation of velocity of sound,} \\ v=\sqrt{\dfrac{Y}{\rho }}.............\text{ }(1) \\ \text{The young }\!\!'\!\!\text{ s modulus of elasticity,} \\ Y=\dfrac{stress}{strain} \\ \text{Substituting the values,} \\ Y=\dfrac{300}{60} \\ \Rightarrow Y=5\text{ }N{{m}^{-2}} \\ \text{Substituting the values in equation (1),} \\ v=\sqrt{\dfrac{Y}{\rho }} \\ \Rightarrow 331=\sqrt{\dfrac{5}{\rho }} \\ \text{Rearranging the equation,} \\ \rho =\dfrac{331\times 331}{5} \\ \therefore \rho =21912.2\text{ }kg{{m}^{-3}}$

 Let us assume that there are $\text{3}$ nodes and $\text{3}$ antinodes when the frequency of the wave inside a closed organ pipe is $500\text{ }Hz$. Find out the length of the organ pipe. 

 The wavelength of a wave will be the ratio of the velocity of the wave to the frequency of the wave. Substitute this in the equation of length of closed pipe. 

 $\text{The equation of length of the organ pipe if there are 3 nodes and 3 antinodes,} \\ L=\dfrac{\left( 2n-1 \right)\lambda }{4}.............\text{ }(1) \\ \text{The wavelength,} \\ \lambda =\dfrac{c}{\nu } \\ \text{Substituting the values,} \\ \lambda =\dfrac{3\times {{10}^{8}}}{500} \\ \Rightarrow \lambda =6\times {{10}^{5}}\text{ }m \\ \text{Substituting the values in equation (1),} \\ L=\dfrac{\left( 2\times 3-1 \right)\times 6\times {{10}^{5}}}{4} \\ \Rightarrow L=\dfrac{30\times {{10}^{5}}}{4} \\ \therefore L=7.5\times {{10}^{5}}\text{ }m$

 Aman, took a $10\;m$ long string to which he attached  a weight of  $5\;kg$, he then gently puckered the strings to  produce waves. Calculate the velocity at which the waves will travel. 

 When a weight is attached to a string it acts as a stretched string, and when this stretched string is disturbed it produces transverse waves over it. 

 $\text{We know that,} \\ v=\sqrt{\dfrac{T}{\mu }} \\ \text{Where}~T=mg\text{ is the tension on the string}. \\ \text{And} \\ \mu =\dfrac{m}{l} \\ \text{Then, we have,} \\ T=5\times 10=50\text{ }N \\ \text{And} \\ \mu =\dfrac{5}{10}=0.5\text{ }kg{{m}^{-1}} \\ \text{Substituting the values, we have} \\ v=\dfrac{50}{0.5}=100\text{ }m{{s}^{-1}}$