Factors on which intensity depends are
(i) ) Amplitude $$\left( a \right)$$ of vibration of the source, $$I \propto {a^2}$$
(ii) Surface area $$\left( A \right)$$ of the vibrating body, $$I \propto A$$
(iii) Density $$\left( \rho \right)$$ of the medium, $$I \propto \rho $$
(iv) Frequency $$\left( \nu \right)$$ of the source, $$I \propto {\nu ^2}$$
(v) Motion of the medium which changes effective velocity $$v$$ of sound, $$I \propto v$$
$${\text{As}}\,\,I \propto {a^2}\,\,{\text{and}}\,\,I \propto {\nu ^2}$$
Therefore, intensity becomes $$\frac{{{2^2}}}{{{4^2}}} = \frac{1}{4}th.$$
32.
A progressive sound wave of frequency $$500\,Hz$$ is travelling through air with a speed of $$350\,m{s^{ - 1}}.$$ A compression maximum appears at a place at a given instant. The minimum time interval after which the rare fraction maximum occurs at the same point, is
33.
Equation of a progressive wave is given by
$$y = 4\sin \left[ {\pi \left( {\frac{t}{5} - \frac{x}{9}} \right) + \frac{\pi }{6}} \right]$$
Then which of the following is correct ?
The standard equation of a progressive wave is
$$y = a\sin \left[ {2\pi \left( {\frac{t}{T} - \frac{x}{\lambda }} \right) + \phi } \right]$$
The given equation can be written as
$$\eqalign{
& y = 4\sin \left[ {2\pi \left( {\frac{t}{{10}} - \frac{x}{{18}}} \right) + \frac{\pi }{6}} \right] \cr
& \therefore a = 4\,cm,T = 10\,s,\lambda = 18\,cm\,{\text{and}}\,\phi = \frac{\pi }{6} \cr} $$
Hence, (B) is correct.
34.
A police car moving at $$22\,m/s,$$ chases a motorcyclist. The police man sounds his horn at $$176\,Hz,$$ while both of them move towards a stationary siren of frequency $$165\,Hz.$$ Calculate the speed of the motorcycle, if it is given that he does not observes any beats.
$${f_1} = $$ frequency of the police car heard by motorcyclist,
$${f_2} = $$ frequency of the siren heard by motorcyclist.
$$\eqalign{
& {f_1} = \frac{{330 - v}}{{330 - 22}} \times 176;{f_2} = \frac{{330 + v}}{{330}} \times 165; \cr
& \because \,\,{f_1} - {f_2} = 0 \cr
& \Rightarrow \,\,v = 22\,m/s \cr} $$
35.
A wave travelling along the $$x$$ - axis is described by the equation $$y\left( {x,t} \right) = 0.005\cos \left( {\alpha x - \beta t} \right).$$ If the wavelength and the time period of the wave are $$0.08\,m$$ and $$2.0\,s,$$ respectively, then $$\alpha $$ and $$\beta $$ in appropriate units are
A
$$\alpha = 25.00\,\pi ,\beta = \pi $$
B
$$\alpha = \frac{{0.08}}{\pi },\beta = \frac{{2.0}}{\pi }$$
C
$$\alpha = \frac{{0.04}}{\pi },\beta = \frac{{1.0}}{\pi }$$
D
$$\alpha = 12.50\,\pi ,\beta = \frac{\pi }{{2.0}}$$
36.
A cylindrical resonance tube open at both ends, has a fundamental frequency $$f,$$ in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be
Fundamental frequency of open pipe,
\[f = \frac{v}{{2l}}\,.......\left( {\text{i}} \right)\,\,\left[ {\begin{array}{*{20}{c}}
{v = {\text{velocity of wave}}} \\
{l = {\text{length of open pipe}}}
\end{array}} \right]\]
When half length of tube is dipped vertically in water, then length of the air column becomes half $$\left( {l' = \frac{l}{2}} \right)$$ and the pipe becomes closed.
So, new fundamental frequency of closed pipe
$$f' = \frac{v}{{4l}} = \frac{v}{{4\left( {\frac{l}{2}} \right)}} = \frac{v}{{2l}}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get,
$$f' = f$$
Hence, there will be no change in fundamental frequency.
37.
The two nearest harmonics of a tube closed at one end and open at other end are $$220\,Hz$$ and $$260\,Hz.$$ What is the fundamental frequency of the system?
Frequency of nth harmonic in a closed end tube
$$ \Rightarrow f = \frac{{\left( {2n - 1} \right)v}}{{4l}}\,\,n = 1,2,3,\,....$$
Also, only odd harmonics exists in a closed end tube.
Now, given two nearest harmonics are of frequency $$220\,Hz$$ and $$260\,Hz.$$
$$\therefore \frac{{\left( {2n - 1} \right)v}}{{4l}} = 220\,Hz\,......\left( {\text{i}} \right)$$
Next harmonic occurs at
$$\frac{{\left( {2n + 1} \right)v}}{{4l}} = 260\,Hz\,......\left( {{\text{ii}}} \right)$$
On subtracting Eq. (i) from Eq. (ii), we get
$$\eqalign{
& \frac{{\left\{ {\left( {2n + 1} \right) - \left( {2n - 1} \right)} \right\}v}}{{4l}} = 260 - 220 \cr
& 2\left( {\frac{v}{{4l}}} \right) = 40 \Rightarrow \frac{v}{{4l}} = 20\,Hz \cr} $$
$$\therefore $$ Fundamental frequency of the system
$$ = \frac{v}{{4l}} = 20\,Hz$$
38.
Standing waves are produced in a $$10\,m$$ long stretched string. If the string vibrates in 5 segments and the wave velocity is $$20\,m/s,$$ the frequency is
In the case of standing wave, the length of one segment is $$\frac{\lambda }{2}.$$ There are 5 segments and total length of string is $$10\,m.$$
$$\eqalign{
& \therefore 5\frac{\lambda }{2} = 10 \cr
& \Rightarrow \lambda = 4\,cm \cr
& {\text{Frequency,}}\,\,n = \frac{v}{\lambda } = \frac{{20}}{4} = 5\,Hz\,\,\left( {\because v = 20\,m/s} \right) \cr} $$ NOTE
Standing wave is an example of interference. Destructive interference means node and constructive interference means antinode.
39.
A speeding motorcyclist sees traffic jam ahead of him. He slows down to $$36\,km/hour.$$ He finds that traffic has eased and a car moving ahead of him at $$18\,km/hour$$ is honking at a frequency of $$1392\,Hz.$$ If the speeds of sound is $$343\,m/s,$$ the frequency of the honk as heard by him will be :
According to Doppler's effect
Apparent frequency
$$\eqalign{
& n' = n\left( {\frac{{v + {v_0}}}{{v + {v_s}}}} \right) = 1392\left( {\frac{{343 + 10}}{{343 + 5}}} \right) \cr
& = 1412\,Hz \cr} $$
40.
A granite rod of $$60\,cm$$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $$2.7 \times {10^3}kg/{m^3}$$ and its Young's modulus is $$9.27 \times {10^{10}}Pa.$$ What will be the fundamental frequency of the longitudinal
vibrations?
