Numerical Problems-1 - Chapter 11 Waves 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
The speed of a wave in a certain medium is $900 \mathrm{~m} / \mathrm{s}$. If 3000 waves passes over a certain point of the medium in 2 minutes, then compute its wavelength?
Answer:
Speed of the wave in medium $\mathrm{v}=900 \mathrm{~ms}^{-1}$
Frequency $(n)=\frac{\text { Number of waves }}{\text { Time }}=\frac{3000}{2 \times 60}=25 \mathrm{~s}^{-1}$
Wavelength $\lambda=\frac{v}{n}=\frac{900}{25} ; \lambda=36 \mathrm{~m}$
Question 2.
Consider a mixture of $2 \mathrm{~mol}$ of helium and $4 \mathrm{~mol}$ of oxygen. Compute the speed of sound in this gas mixture at $300 \mathrm{~K}$.
Answer:
The mixture of helium and oxygen.

$
\begin{aligned}
& \mathrm{M}_{\text {mix }}=\frac{n_1 \mathrm{M}_1+n_2 \mathrm{M}_2}{n_1+n_2}=\frac{(2 \times 4)+(4 \times 32)}{2+4}=\frac{136}{6} \\
& \mathrm{M}_{\text {mix }}=22.6 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}
\end{aligned}
$
Helium is an mono atomic, $\mathrm{C}_{\mathrm{v}_1}$, $=\frac{3 \mathrm{R}}{2}$
Oxygen is an diatomic, $\mathrm{C}_{\mathrm{v}_2}=\frac{5 \mathrm{R}}{2}$
$
\begin{aligned}
\left(\mathrm{C}_{\mathrm{v}}\right)_{\text {mix }} & =\frac{n_1 \mathrm{C}_{\mathrm{v}_1}+n_2 \mathrm{C}_{\mathrm{v}_2}}{n_1+n_2}=\frac{\left(2 \times \frac{3 \mathrm{R}}{2}\right)+\left(4 \times \frac{5 \mathrm{R}}{2}\right)}{2+4} \\
\left(\mathrm{C}_{\mathrm{v}}\right)_{\text {mix }} & =\frac{13 \mathrm{R}}{6} \\
\left(\mathrm{C}_{\mathrm{p}}\right)_{\text {mix }} & =\left(\mathrm{C}_{\mathrm{v}}\right)_{\text {mix }}+\mathrm{R} \quad \gamma_{\text {mix }}=\frac{\left(\mathrm{C}_{\mathrm{P}}\right)_{\text {mix }}}{\left(\mathrm{C}_{\mathrm{V}}\right)_{\text {mix }}}=\frac{19 \mathrm{R}}{6} \times \frac{6}{13 \mathrm{R}} \\
& =\frac{13 \mathrm{R}}{6}+\mathrm{R}=\frac{19 \mathrm{R}}{6} \quad \gamma_{\text {mix }}=\frac{19}{13}
\end{aligned}
$
According to Laplace correction, Speed of sound,
Extend root length litre this $v=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{19}{13} \times \frac{8.31 \times 300}{22.6 \times 10^{-3}}} ; v=400.9 \mathrm{~ms}$

Question 3.
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after $3.5 \mathrm{~s}$. After the ship moves to $100 \mathrm{~km}$ it sends another signal which returns back after 2 s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.
Answer:
Speed of SONAR waves in water $\mathrm{c}=1500 \mathrm{~ms}^{-1}$
Time taken to reflect from the bottom of the sea, $2 \mathrm{t}=3.5 \mathrm{sec}$
$\therefore \mathrm{t}=1.75 \mathrm{sec}$
Distance covered in forward and reflected backward $\left(\mathrm{d}_1\right)=\mathrm{c} \times \mathrm{t}$
$\mathrm{d}_2=1500 \times 1.75=2625 \mathrm{~m}$
After ship moves in a distance $=150 \mathrm{~km}$
Time taken to reflect by the waves $2 \mathrm{t}=2 \mathrm{~s}$ $\mathrm{t}=1 \mathrm{~s}$
Distance covered by the waves $\left(\mathrm{d}_2\right)=\mathrm{c} \times \mathrm{t}=1500 \times 1=1500 \mathrm{~m}$
The different between the height of two cases $=2625-1500$ $\mathrm{h}_{\text {difference }}=1124 \mathrm{~m}$
Question 4.
A sound wave is transmitted into a tube as shown in figure. The sound wave splits into two waves at the point A which recombine at point $\mathrm{B}$. Let $\mathrm{R}$ be the radius of the semicircle which is varied until the first minimum. Calculate the radius of the semicircle if the wavelength of the sound is $50.0 \mathrm{~m}$
Answer:
The sound travelling in the curved path distance $=\pi \mathrm{R}$
$\mathrm{L}_1=\pi \mathrm{R}$
The sound travelling in the straight path distance $=2 R$
$\mathrm{L}_2=2 \mathrm{R}$

The path distance of straight and curved path $\mathrm{AP}=\mathrm{L}_1-\mathrm{L}_2$
$
\Delta \mathrm{P}=\pi \mathrm{R}-2 \mathrm{R}=\mathrm{R}(\pi-2)
$
The different in the path length of the sound waves,
$
\Delta \mathrm{P}=\frac{\lambda}{2}
$
Equatipg (1) and (2), $\frac{\lambda}{2}=\mathrm{R}(\pi-2) ; \lambda=50 \mathrm{~m}$
$
\begin{aligned}
& \mathrm{R}=\frac{\lambda}{2(\pi-2)}=\frac{50}{2(3.14-2)}=\frac{50}{2.28} \\
& \mathrm{R}=21.9 \mathrm{~m}
\end{aligned}
$
Question 5.
$\mathrm{N}$ tuning forks are arranged in order of increasing frequency and any two successive tuning forks give $n$ beats per second when sounded together. If the last fork gives double the frequency of the first (called as octave), Show that the frequency of the first tuning fork is $f=(N-1) n$.
Answer:
Total number of fork $=\mathrm{N}$
The frequency of the 1 st fork $=f$
The frequency of the last fork $=2 \mathrm{f}$
$
\begin{aligned}
& \therefore \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \\
& 2 \mathrm{f}=\mathrm{f}+(\mathrm{N}-1) \mathrm{n} \\
& 2 \mathrm{f}-\mathrm{f}=(\mathrm{N}-1) \mathrm{n} \\
& \therefore \mathrm{f}=(\mathrm{N}-1) \mathrm{n}
\end{aligned}
$

Question 6.
Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?
Answer:
Intensity of sound wave (old) $=\mathrm{I}_1$
Amplitude of sound wave $\left(\mathrm{A}_2\right)=2 \mathrm{~A}_1$
Frequency of the sound wave $I_2=$ ?
$\mathrm{I}_1 \propto f_1^2 \mathrm{~A}_1^2 ; \mathrm{I}_2 \propto f_2^2 \mathrm{~A}_2^2$
$
\begin{aligned}
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{f_1^2 \cdot \mathrm{A}_1^2}{f_2^2 \cdot \mathrm{A}_2^2}=\frac{f_1^2 \mathrm{~A}_1^2}{\frac{1}{16} f_1^2 \cdot 4 \mathrm{~A}_1^2}=\frac{16}{4}=4 \\
& \mathrm{I}_2=\frac{1}{4} \mathrm{I}_1
\end{aligned}
$
Question 7.
Consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is $250 \mathrm{~Hz}$. Calculate the fundamental frequency of the open pipe.
Answer:
Fundamental frequency of closed organ pipe
$
f_c=\frac{v}{4 l}=250 \mathrm{~Hz}
$

Fundamental frequency:f open organ pipe $f_o=\frac{v}{2 l}=?$
$
\begin{aligned}
& \frac{f_c}{f_o}=\frac{v}{4 l} \times \frac{2 l}{v}=\frac{1}{2} \\
& f_o=2 f_c=2 \times 250 \\
& f_o=500 \mathrm{~Hz}
\end{aligned}
$
Question 8.
A police in a siren car moving with a velocity $20 \mathrm{~ms}^{-}$chases a thief who is moving in a car with a velocity $\mathrm{v}_0 \mathrm{~ms}^{-1}$. The police car sounds at frequency $300 \mathrm{~Hz}$, and both of them move towards a stationary siren of frequency $400 \mathrm{~Hz}$. Calculate the speed in which thief is moving.
(Assume the thief does not observe any beat)
Answer:
Velocity of sound $\mathrm{v}=330 \mathrm{~ms}^{-1}$
Velocity of car $\left(\mathrm{v}_{\mathrm{s}}\right)=20 \mathrm{~ms}^{-1}$
Frequency of $\operatorname{car}\left(f_1\right)=300 \mathrm{~Hz}$
Frequency of stationary siren $\left(f_2\right)=400 \mathrm{~Hz}$
The speed of the thief $\left(\mathrm{v}_{\mathrm{o}}\right)=$ ?
$
v_1=f_1\left[\frac{v-v_o}{v-v_s}\right] \text { and } v_2=f_2\left[\frac{v+v_o}{v}\right]
$
Both are moving towards stationary $\operatorname{siren} v_1=v_2$
$
\begin{aligned}
300\left[\frac{300-v_o}{310}\right] & =400\left[\frac{330+v_o}{330}\right] \\
330-v_o & =\frac{4 \times 310}{3}\left[\frac{330+v_o}{330}\right] \\
330-v_o & =\frac{1240}{990}\left[330+v_0\right]=1.2525\left(330+v_0\right)
\end{aligned}
$

$
\begin{aligned}
& 330-\mathrm{v}_{\mathrm{o}}=413.325+1.2525 \mathrm{v}_{\mathrm{o}} \\
& 2.2525 \mathrm{v}_{\mathrm{o}}=-83.325
\end{aligned}
$
$
\mathrm{v}_{\mathrm{o}}=-36.99
$
$\therefore$ speed of the thief in moving $=36.99 \mathrm{~ms}^{-1}$
Question 9.
Consider the following function:
(a) $y=x^2+2 \alpha t x$
(b) $y=(x+v t)^2$ which among the above function can be characterized as a wave?
Answer:
(a) $y=x^2+2 \alpha \mathrm{tx}$. This expression is not a wave equation.
(b) $y=(x+v t)^2$. This expression is satisfies the wave equation.