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This is second of our many quizzes on Standing waves. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Standing waves Quiz 2

physics

 Consider a sphere of radius $\left( {{r}_{0}} \right)$ which emits power $({{p}_{0}})$ uniformly through the space at a distance of $\left( r \right)$ from the source. Find the wave's amplitude in relation to $\left( r \right)$? 

 $I=\dfrac{P}{Area}$, where $I$ is the intensity, $P$ is power and area of the place where Intensity is being measured. Also, $I\propto {{A}^{2}}$, where $A$ is the wave's amplitude.

  $I=\dfrac{{{P}_{0}}}{Area} \\ \text{Total surface area of the sphere =}4\pi {{r}^{2}} \\I=\dfrac{{{P}_{0}}}{4\pi {{r}^{2}}} \\ \text{As I}\propto {{A}^{2}}\\ \Rightarrow {{A}^{2}}\propto \dfrac{1}{{{r}^{2}}} \\ \therefore A\propto \dfrac{1}{r}$

 A guitarist strikes a guitar string to perform the ${{5}^{th}}$ harmonic in order to tune the guitar. If the string's length is $\dfrac{1}{4}\,m,$ its mass is $100\,g$ and assume that the tension in the string is $T=250\,N,$ then calculate the frequency of the played note.

The speed of a transverse wave is $V=\sqrt{\dfrac{T}{\mu },}$ $T=$ Tension in string, $\mu =$ Linear mass density. Then use the formula of ${{5}^{th}}$ harmonic ${{\nu }_{5}}=5\left( \dfrac{V}{4L} \right)$. 

  Let the velocity of the wave be $V$. Then for a transverse wave, $ \\V=\sqrt{\dfrac{T}{\mu }},\mu =\text{Linear mass density} \\ \Rightarrow V=\sqrt{\dfrac{250}{\dfrac{M}{L}}}=\sqrt{\dfrac{250\times 1000}{100\times 4}} \\ \Rightarrow V=25m/s \\ $ To play the fifth harmonic, the frequency will be $ \\ \Rightarrow {{\nu }_{5}}=5\left( \dfrac{V }{2L} \right) \\ \Rightarrow {{\nu }_{5}}=\dfrac{5\times 25}{2\times \dfrac{1}{4}} \\ \therefore {{\nu }_{5}}=250\,Hz$

A $300Hz$ fundamental frequency is played by an open organ pipe at one end. The pipe length measures $50cm$. What is the velocity of sound?

 The organ pipe, which is open at one end, has ${{\dfrac{1}{4}}^{th}}$ of the wave in its length for the first harmonic.So its fundamental frequency is $\nu =\dfrac{V}{4L}$. For the 3rd Harmonic, the frequency would be ${{\nu }_{3}}=3\left( \dfrac{V}{4L} \right)$.

 $\text{For an organ pipe open at one end, the fundamental frequency is } \\ \nu =\dfrac{V}{4L},\text{ where V is velocity of sound}\text{.} \\ \text{For }{{\text{3}}^{\text{rd}}}\text{ Harmonic, the frequency will be} \\{{\nu }_{3}}=3\left( \dfrac{V}{4L} \right) \\ \Rightarrow 900=3\left( \dfrac{V}{4\left( \dfrac{1}{2} \right)} \right) \\ \therefore \nu =600\text{ m/s}$

 A car with a siren on top is driving at a speed of $10\text{m/s}$. There is a man who is listening to the sound of a siren and receives a $20\%$ more frequency. Assuming that the man is moving, then determine its speed $\left( {{V}_{sound}}=340\text{m/s} \right)$ 

Use doppler’s effect formula ${{f}_{o}}=\left( \dfrac{V+{{V}_{o}}}{V-{{V}_{s}}} \right){{f}_{s}}$ where ${{f}_{o}}=$ observed frequency of sound, ${{f}_{s}}=$ actual frequency of sound, ${{V}_{o}}=$ observer velocity, ${{V}_{s}}=$ source velocity, $V=$ velocity of sound.

 $f={{f}_{0}}\left( \dfrac{v+{{v}_{0}}}{v-{{v}_{3}}} \right) \\ $ If source moving towards observer then take $-{{v}_{3}}$ and vice versa is true $ \\ $ If observes moving towards the source then take $+{{v}_{0}}$  and vice versa is true $ \\1.2{{f}_{0}}={{f}_{0}}\left( \dfrac{v+{{v}_{0}}}{v-10} \right) \\ \Rightarrow \left( 1.2v-v \right)-12={{v}_{0}} \\ \Rightarrow {{v}_{0}}=0.2\times 340-12 \\ \therefore {{v}_{0}}=68-12=56\,m/s$

 What is the speed of sound at room temperature, i.e. ${{27}^{\circ }}C$ $\left( 300K \right)$. Provided that the speed of sound is $273\,K$ at $330\,m/s$. 

 To calculate the velocity $V$ of sound, use the following formula: $=V={{V}_{o}}\left( 1+\dfrac{t}{546} \right)$ where ${{V}_{o}}=330\text{m/s}$ or speed of sound at $273K,\text{ }t=$ temperature in ${}^\circ C$. 

 $\text{Velocity of sound at given temperature } t{}^\circ C is \\ V={{V}_{0}}\left( 1+\dfrac{t}{546} \right) \\{{\text{V}}_{o}}=330\text{ }m/s,\text{ t=27  }\!\!{}^\circ\!\!\text{ }C \\V={{V}_{0}}+\dfrac{{{V}_{0}}}{546}t=330+\dfrac{330}{546}\times 27 \\ \Rightarrow V= 330+0.6\times 27 \\ \therefore V\approx 346.2\,m/s$ 