Numerical Problems-2 - Chapter 11 - Waves - 11th Science Guide Samacheer Kalvi Solutions

You can Download the Numerical Problems-2 - Chapter 11 - Waves - 11th Science Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Numerical Problems
Question 1.
The fundamental frequency in an open organ pipe is equal to the 3rd harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \mathrm{~cm}$. What is the length of the open organ pipe.
Answer:
For closed organ pipe, 3 rd harmonics $=\frac{3 v}{4 l}$
For open organ pipe, fundamental frequency, $=\frac{v}{2 l^{\prime}}$

Given,
$
l=20 \mathrm{~cm}
$
$
\begin{aligned}
\frac{3 v}{4 l} & =\frac{v}{2 l^{\prime}} \\
l^{\prime} & =\frac{4 l}{3 \times 2}=\frac{4 l}{6} \\
l^{\prime} & =\frac{4 \times 20}{6} \\
l^{\prime} & =13.33 \mathrm{~cm}
\end{aligned}
$
Question 2.
Two cars moving in opposite directions approach each other with speed of $22 \mathrm{~ms}^{-1}$ and $16.5 \mathrm{~ms}^{-1}$ respectively. The driver of the first car blows a horn having a frequency 400 $\mathrm{Hz}$. To find the frequency heard by the driver of the second car.
Answer:
$
\begin{aligned}
f_{\mathrm{A}} & =f\left[\frac{v+v_o}{v-v_s}\right]=400\left[\frac{340+16.5}{340-22}\right] \\
& =400\left[\frac{356.5}{318}\right]=400 \times 1.1210 \\
f_{\mathrm{A}} & =448.4 \mathrm{~Hz}
\end{aligned}
$
Question 3.
The second overtone of an open organ pipe has the same frequency as the 1 st overtone of a closed pipe L metre long. Then what will be the length of the open pipe.
Answer:

$2^{\text {nd }}$ overtone of an open organ pipe $=\frac{3 v}{2 \mathrm{~L}_{\mathrm{O}}}$
$
\begin{aligned}
\frac{3 v}{3 \mathrm{~L}_{\mathrm{O}}} & =\frac{3 v}{4 \mathrm{~L}_{\mathrm{C}}} \\
\frac{1}{\mathrm{~L}_{\mathrm{O}}} & =\frac{1}{2 \mathrm{~L}_{\mathrm{C}}} \\
\mathrm{L}_{\mathrm{O}} & =2 \mathrm{~L}_{\mathrm{C}}
\end{aligned}
$
The Length of the open pipe is two times of the length of the closed pipe.
Question 4.
A steel wire $0.72 \mathrm{~m}$ long has a mass of $5.0 \times 10^{-3} \mathrm{~kg}$. If the wire is under a tension of 60 $\mathrm{N}$. What is the speed of transverse waves on the wire?
Answer:
Here $\mathrm{T}=60 \mathrm{~N}$, Mass $=5.0 \times 10^{-3} \mathrm{Kg}$, Length $=0.72 \mathrm{~m}$
mass per unit length, $\quad m=\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}=6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$
The speed of the transverse wave on the wire
$
v=\sqrt{\frac{T}{m}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}} \Rightarrow v=93 \mathrm{~ms}^{-1}
$
Question 5.
Estimate the speed of sound $; \mathrm{n}$ air at standard temperature and pressure by using
(i) Newton's formula and
(ii) Laplace formula. The mass of 1 moIe of air $=29.0 \times 10^{-3} \mathrm{~kg}$. For air, $\gamma=1.4$

Answer:

Density of air,
$
\begin{aligned}
\rho & =\frac{\text { Mass of } 1 \text { mole of air }}{\text { Volume of } 1 \text { mole of air }} \\
& =\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.9 \text { litre }}=\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}=1.29 \mathrm{~kg} \mathrm{~m}^{-3}
\end{aligned}
$
Standard pressure, $\quad \mathrm{P}=1.01 \times 10^5 \mathrm{~Pa}$.
(i) According to Newton's formula, speed of sound in air at S.T.P is
$
v=\sqrt{\frac{\mathrm{P}}{\rho}}=\sqrt{\frac{1.01 \times 10^5}{1.29}}=280 \mathrm{~ms}^{-1}
$
(ii) According to Newton's formula speedof sound in air at S.T.P is
$
v=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}=\sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}}=331.5 \mathrm{~ms}^{-1}
$
Question 6.
An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What Is the percentage increase in the apparent frequency?
Answer:
Here observer moves towards the stationary source.
$
v_0=-v / 5, v_s=0
$
Apparent frequency $f^{\prime}=\frac{v-v_o}{v-v_s} \times f=\frac{v-v / 5}{v-0} \times f=\frac{6}{5} v=1.2 v$
The percentage increase in apparent frequency
$
\frac{f^{\prime}-f}{f} \times 100=\frac{1.2 f-f}{f} \times 100=20 \%
$
Question 7.
Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What Is the ratio of fundamental frequency of the tubes A and B?
Answer:
The fundamental frequency for tube $\mathrm{A}$ with both ends open is $\mathrm{f}_{\mathrm{A}}=\frac{v}{2 \mathrm{~L}}$
The fundamental frequency for tube $B$ with one end closed is $\mathrm{f}_{\mathrm{B}}=\frac{v}{4 \mathrm{~L}}$

$
\frac{f_{\mathrm{A}}}{f_{\mathrm{B}}}=\frac{v / 2 \mathrm{~L}}{v / 4 \mathrm{~L}}=2 \Rightarrow f_{\mathrm{A}}: f_{\mathrm{B}}=2: 1
$
Question 8.
A train moves towards a stationary observer with speed $34 \mathrm{mIs}$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the train's speed is reduced to $17 \mathrm{~m} / \mathrm{s}$, the frequency registered $f_2$. If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, then find the ratio $\mathrm{f}_1 / \mathrm{f}_2$
Answer:
For the stationary observer $f_1=\frac{v}{v-v_s} \times f$
Hence
$
\begin{aligned}
& \therefore \quad f_1=\frac{340}{340-34} \times f \text { and } f_2=\frac{340}{340-17} \times f \\
& \frac{f_2}{f_1}=\frac{340-17}{340-34}=\frac{19}{18} \Rightarrow f_1: f_2=19: 18
\end{aligned}
$
Question 9.
A police car with a siren of frequency $8 \mathrm{kHz}$;s moving with uniform velocity $36 \mathrm{~km} / \mathrm{h}$ towards a tall building which reflect the sound waves. The speed of sound In air is $320 \mathrm{~m} / \mathrm{s}$. What is the frequency of the siren heard by the car driver?
Answer:
(a) Frequency received by the building.
$
f^{\prime}=\left(\frac{v}{v-v_c}\right)
$

The wall (source) reflect this frequency, So frequency heard by the car driver is
$
\begin{aligned}
f^{\prime \prime} & =\left(\frac{v+v_c}{v}\right) f^{\prime}=\left(\frac{v+v_c}{v}\right)\left(\frac{v}{v-v_c}\right) f \\
& =\left(\frac{v+v_c}{v-v_c}\right) f=\left(\frac{320+10}{320-10}\right) \times 8 \mathrm{kHz}=\frac{33}{31} \times 8=8.5 \mathrm{kHz}
\end{aligned}
$
Question 10.
The displacement $\mathrm{y}$ of a wave travelling in the $\mathrm{x}$-direction $;$ s given by $\mathrm{y}=10^{-4} \sin (600 \mathrm{t}$ $-2 \mathrm{x}+\pi / 3)$
Where $\mathrm{x}$ is expressed in metres and $\mathrm{t}$ is seconds. What is the speed of the wave motion (in $\mathrm{ms}^{-1}$ ?
Answer:
$
\begin{aligned}
& y=10^{-4} \sin \left(600 t-2 x+\frac{\pi}{3}\right) \text { and } y=a \sin \left(\omega t-k x+\frac{\pi}{3}\right) \\
& \therefore \quad \omega=600 \mathrm{rad} \mathrm{s}^{-1}, k=2 \mathrm{rad}^{-1} \mathrm{~m} \\
& \nu=\frac{\omega}{k}=\frac{600}{2}=300 \mathrm{~ms}^{-1} \\
&
\end{aligned}
$