Numerical Problems-2 - Chapter 11 Waves 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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}
$