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This is one of our many quizzes on Frequency and pitch. Click on this to access more such competitive quizzes on Waves.

Frequency and pitch Quiz 2

physics

 In which of the following situations, you would not observe a difference in frequencies between the actual and the apparent?

  Difference of frequencies Is observed only when there exists relative motion between source and observer.

  If the distance between observer and source is not constant, that means there exists relative velocity between both of them, and hence, it means that there exists a difference of frequencies. If only the medium is moving, that means there is no relative motion between observer and source. Hence, no difference exists between apparent and actual frequencies. Two objects moving in a mutually perpendicular direction do not feel any difference.

 A bus moving with a speed of $36.6\,m{{s}^{-1}}$ produces a horn of a certain frequency. How much apparent frequency would be felt by a student when the bus crosses him?

 Recall the Doppler’s effect to calculate the apparent frequency when the object is coming towards the observer and when the object is going away from the observer.

${f}'=\dfrac{{{f}_{v}}}{v-{{v}_{s}}} \\ {f}''=\dfrac{{{f}_{v}}}{v+{{v}_{s}}} \\ \Rightarrow \dfrac{{{f}''}}{{{f}'}}=\dfrac{v-{{v}_{s}}}{v+{{v}_{s}}} $ Given that ${{v}_{s}}$= $36.6\,m{{s}^{-1}}$ and the speed of sound in air, $v=330\,m{{s}^{-1}}$. Substituting the values, we get $\dfrac{{{f}''}}{{{f}'}}=\dfrac{4}{5}$ Hence, the student will feel ${{\left( \dfrac{4}{5} \right)}^{th}}$ of the initial frequency when the bus passes him.

 A string of guitar of length $65\,cm$ has mass $20\,g$ is stretched such that tension in the string becomes $500\,N$ . If it produces a sound wave of wavelength$0.0002\,m$, then calculate the power of the wave produced for $t=5\text{ }s$.

 Calculate the velocity of the sound wave and hence, find the frequency of the sound wave. Using this, find the energy of the wave and then power by dividing energy to the time.

&nbsp;First of all, we need to calculate the velocity of the sound wave For stretched string, the formula to calculate velocity of sound wave is given by - $v=\sqrt{\dfrac{T}{m}} \\ \Rightarrow m=\dfrac{0.002}{0.65}=0.003 \\ \Rightarrow v=\sqrt{\dfrac{500}{0.003}}=1290.9m{{s}^{-1}} $ Here, $m$ is the linear density of the string. Frequency = $\dfrac{v}{\lambda }=\dfrac{1290.9}{0.2}=6,454.5\,Hz$ Hence, the required frequency is $11,180\,Hz$. Therefore the energy of the wave, $E=h\nu $ Substituting the values in it, $E=6.626\times {{10}^{-34}}\times 11180 \\ \Rightarrow E=74078.68\times {{10}^{-34}} \\ \Rightarrow E=7.41\times {{10}^{-30}}\text{ }J $ The power can be found, $E=Pt \\ \Rightarrow P=\dfrac{E}{t} $ Substituting the values in it, $P=\dfrac{7.41\times {{10}^{-30}}}{5} \\ \therefore P=1.482\text{ }W$

 Determine the kinetic energy and the momentum of the sound wave, having a wavelength of $0.006\,m$ and is moving in certain gas which has molecular weight $2\,g$ and $\gamma $ value is 1.4.Given, the temperature is ${{0}^{o}}C$. 

  Calculate the speed of sound in the given gas and substitute the value of calculated velocity of sound in the general formula of frequency to get the correct answer.

Calculating the Speed of the sound in the given gas – $\dfrac{P}{\rho }=\dfrac{RT}{M}=\dfrac{8.3\times 273}{2\times {{10}^{-3}}} \\ \Rightarrow \dfrac{P}{\rho }=11.34\times {{10}^{5}} $ Now, $v=\sqrt{\dfrac{\gamma P}{\rho }} \\ \Rightarrow v=\sqrt{1.4\times 11.34\times {{10}^{5}}} \\ \Rightarrow v=1260m{{s}^{-1}}$ The equation of kinetic energy, $E=\dfrac{1}{2}M{{v}^{2}}$ Substituting the values in it, $E=\dfrac{1}{2}\times 2\times {{10}^{-3}}\times {{\left( 1260 \right)}^{2}} \\ \therefore E=1.58\times {{10}^{3}}\text{ }J $ The equation of momentum, $P=Mu$ Substituting the values, $P=2\times {{10}^{-3}}\times 1260 \\ \therefore P=2.52\text{ }W$

 Consider a source which is emitting a sound of frequency $50\text{ }Hz$, placed at a large distance from the observer. If the source starts moving towards the observer with an acceleration $5\text{ }m{{s}^{-2}}$ and the speed of sound in the medium is $40\text{ }m{{s}^{-1}}$ , then the apparent frequency is given by – 

 First of all, calculate the apparent change in wavelength of the source. Substitute the obtained expression in the formula of frequency to get the expression for the apparent frequency.

Since the source starts moving with uniform acceleration towards the observer, the wavelength of the waves heard by the observer decreases by $\dfrac{1}{2}a{{T}^{2}}$. Hence, the new wavelength becomes – ${\lambda }'=\lambda -\dfrac{1}{2}a{{T}^{2}}$ Therefore, the apparent frequency ${f}'=\dfrac{v}{{{\lambda }'}}=\dfrac{v}{~\lambda -\dfrac{1}{2}a{{T}^{2}}} \\ \text{Now, since T is the time period, } \\ \text{put }T=\dfrac{1}{f} \\ \Rightarrow {f}'=\dfrac{v}{~\lambda -\dfrac{1}{2}a{{\left( \dfrac{1}{f} \right)}^{2}}} \\ \Rightarrow {f}'=\dfrac{2{{f}^{2}}v}{2fv-a} $ Substituting the values in it, ${f}'=\dfrac{2\times {{\left( 50 \right)}^{2}}\times 40}{2\times 50\times 40-5} \\ \Rightarrow {f}'=\dfrac{2\times {{10}^{5}}}{3995} \\ \therefore {f}'=50.06\text{ }Hz $ Hence, the apparent frequency is $50.06\text{ }Hz$.