173149
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will
1 increase by a factor of 2
2 decrease by a factor of 2
3 decrease by a factor of 4
4 remains unchanged
Explanation:
C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 . We know that, Intensity of sound $(\mathrm{I})=\mathrm{A}^{2} \mathrm{n}^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{2}$ $=\left(\frac{\mathrm{A}}{2 \mathrm{~A}}\right)^{2}\left(\frac{4 \mathrm{n}}{\mathrm{n}}\right)^{2}$ $\mathrm{I}_{2}=2 \mathrm{~A}$ $\mathrm{n}_{1}=4 \mathrm{n}$ $=\frac{1}{4} \times 16=4$ $\therefore \quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{4}$
AIPMT- 1989
WAVES
173150
Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of $40 \mathrm{~cm}$ is equivalent to phase difference of $\mathbf{1 . 6}$ $\pi$. The frequency of this wave is
1 $165 \mathrm{~Hz}$
2 $150 \mathrm{~Hz}$
3 $660 \mathrm{~Hz}$
4 $330 \mathrm{~Hz}$
Explanation:
C Given, Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Path difference $(\Delta \mathrm{x})=40 \mathrm{~cm}=0.4 \mathrm{~m}$ Phase different $(\phi)=1.6 \pi$ We know that, $\frac{\Delta \mathrm{x}}{\Delta \phi}=\frac{\lambda}{2 \pi}$ Put the value in formula, $\frac{0.4}{1.6 \pi}=\frac{\lambda}{2 \pi}$ $\lambda=0.5 \mathrm{~m}$ The frequency of wave, $=\frac{\mathrm{V}}{\lambda}$ $=\frac{330 \mathrm{~m} / \mathrm{s}}{0.5}$ $=660 \mathrm{~Hz}$
AIPMT-1990
WAVES
173151
In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 \mathrm{~s}$. The frequency of the wave is
1 $1.47 \mathrm{~Hz}$
2 $0.36 \mathrm{~Hz}$
3 $0.73 \mathrm{~Hz}$
4 $2.94 \mathrm{~Hz}$
Explanation:
A Maximum displacement time (t) $=0.170 \mathrm{~s}$ For one vibration, Time period, $(\mathrm{T})=4 \mathrm{t}=4 \times 0.17$ $=0.68 \mathrm{sec}$ Frequency $(f)=\frac{1}{T}$ $=\frac{1}{0.68}=1.47 \mathrm{~Hz}$
AIPMT-1998
WAVES
173152
A pipe closed at one end and open at the other end resonates with a sound of frequency $135 \mathrm{~Hz}$ and also with $165 \mathrm{~Hz}$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is
1 $15 \mathrm{~Hz}$
2 $60 \mathrm{~Hz}$
3 $7.5 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
A Given, The sound frequency $\left(\mathrm{f}_{1}\right)=135 \mathrm{~Hz}$ $\mathrm{f}_{2}=165 \mathrm{~Hz}$ The different between two frequencies, $\mathrm{f}_{2}-\mathrm{f}_{1}=165-135=30 \mathrm{~Hz}$ According to question, $\mathrm{f}_{2}-\mathrm{f}_{1}=2 \times \text { fundamental frequency }$ $30=2 \times \text { fundamental frequency }$ $\therefore$ Fundamental frequency $=15 \mathrm{~Hz}$
173149
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will
1 increase by a factor of 2
2 decrease by a factor of 2
3 decrease by a factor of 4
4 remains unchanged
Explanation:
C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 . We know that, Intensity of sound $(\mathrm{I})=\mathrm{A}^{2} \mathrm{n}^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{2}$ $=\left(\frac{\mathrm{A}}{2 \mathrm{~A}}\right)^{2}\left(\frac{4 \mathrm{n}}{\mathrm{n}}\right)^{2}$ $\mathrm{I}_{2}=2 \mathrm{~A}$ $\mathrm{n}_{1}=4 \mathrm{n}$ $=\frac{1}{4} \times 16=4$ $\therefore \quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{4}$
AIPMT- 1989
WAVES
173150
Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of $40 \mathrm{~cm}$ is equivalent to phase difference of $\mathbf{1 . 6}$ $\pi$. The frequency of this wave is
1 $165 \mathrm{~Hz}$
2 $150 \mathrm{~Hz}$
3 $660 \mathrm{~Hz}$
4 $330 \mathrm{~Hz}$
Explanation:
C Given, Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Path difference $(\Delta \mathrm{x})=40 \mathrm{~cm}=0.4 \mathrm{~m}$ Phase different $(\phi)=1.6 \pi$ We know that, $\frac{\Delta \mathrm{x}}{\Delta \phi}=\frac{\lambda}{2 \pi}$ Put the value in formula, $\frac{0.4}{1.6 \pi}=\frac{\lambda}{2 \pi}$ $\lambda=0.5 \mathrm{~m}$ The frequency of wave, $=\frac{\mathrm{V}}{\lambda}$ $=\frac{330 \mathrm{~m} / \mathrm{s}}{0.5}$ $=660 \mathrm{~Hz}$
AIPMT-1990
WAVES
173151
In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 \mathrm{~s}$. The frequency of the wave is
1 $1.47 \mathrm{~Hz}$
2 $0.36 \mathrm{~Hz}$
3 $0.73 \mathrm{~Hz}$
4 $2.94 \mathrm{~Hz}$
Explanation:
A Maximum displacement time (t) $=0.170 \mathrm{~s}$ For one vibration, Time period, $(\mathrm{T})=4 \mathrm{t}=4 \times 0.17$ $=0.68 \mathrm{sec}$ Frequency $(f)=\frac{1}{T}$ $=\frac{1}{0.68}=1.47 \mathrm{~Hz}$
AIPMT-1998
WAVES
173152
A pipe closed at one end and open at the other end resonates with a sound of frequency $135 \mathrm{~Hz}$ and also with $165 \mathrm{~Hz}$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is
1 $15 \mathrm{~Hz}$
2 $60 \mathrm{~Hz}$
3 $7.5 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
A Given, The sound frequency $\left(\mathrm{f}_{1}\right)=135 \mathrm{~Hz}$ $\mathrm{f}_{2}=165 \mathrm{~Hz}$ The different between two frequencies, $\mathrm{f}_{2}-\mathrm{f}_{1}=165-135=30 \mathrm{~Hz}$ According to question, $\mathrm{f}_{2}-\mathrm{f}_{1}=2 \times \text { fundamental frequency }$ $30=2 \times \text { fundamental frequency }$ $\therefore$ Fundamental frequency $=15 \mathrm{~Hz}$
173149
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will
1 increase by a factor of 2
2 decrease by a factor of 2
3 decrease by a factor of 4
4 remains unchanged
Explanation:
C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 . We know that, Intensity of sound $(\mathrm{I})=\mathrm{A}^{2} \mathrm{n}^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{2}$ $=\left(\frac{\mathrm{A}}{2 \mathrm{~A}}\right)^{2}\left(\frac{4 \mathrm{n}}{\mathrm{n}}\right)^{2}$ $\mathrm{I}_{2}=2 \mathrm{~A}$ $\mathrm{n}_{1}=4 \mathrm{n}$ $=\frac{1}{4} \times 16=4$ $\therefore \quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{4}$
AIPMT- 1989
WAVES
173150
Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of $40 \mathrm{~cm}$ is equivalent to phase difference of $\mathbf{1 . 6}$ $\pi$. The frequency of this wave is
1 $165 \mathrm{~Hz}$
2 $150 \mathrm{~Hz}$
3 $660 \mathrm{~Hz}$
4 $330 \mathrm{~Hz}$
Explanation:
C Given, Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Path difference $(\Delta \mathrm{x})=40 \mathrm{~cm}=0.4 \mathrm{~m}$ Phase different $(\phi)=1.6 \pi$ We know that, $\frac{\Delta \mathrm{x}}{\Delta \phi}=\frac{\lambda}{2 \pi}$ Put the value in formula, $\frac{0.4}{1.6 \pi}=\frac{\lambda}{2 \pi}$ $\lambda=0.5 \mathrm{~m}$ The frequency of wave, $=\frac{\mathrm{V}}{\lambda}$ $=\frac{330 \mathrm{~m} / \mathrm{s}}{0.5}$ $=660 \mathrm{~Hz}$
AIPMT-1990
WAVES
173151
In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 \mathrm{~s}$. The frequency of the wave is
1 $1.47 \mathrm{~Hz}$
2 $0.36 \mathrm{~Hz}$
3 $0.73 \mathrm{~Hz}$
4 $2.94 \mathrm{~Hz}$
Explanation:
A Maximum displacement time (t) $=0.170 \mathrm{~s}$ For one vibration, Time period, $(\mathrm{T})=4 \mathrm{t}=4 \times 0.17$ $=0.68 \mathrm{sec}$ Frequency $(f)=\frac{1}{T}$ $=\frac{1}{0.68}=1.47 \mathrm{~Hz}$
AIPMT-1998
WAVES
173152
A pipe closed at one end and open at the other end resonates with a sound of frequency $135 \mathrm{~Hz}$ and also with $165 \mathrm{~Hz}$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is
1 $15 \mathrm{~Hz}$
2 $60 \mathrm{~Hz}$
3 $7.5 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
A Given, The sound frequency $\left(\mathrm{f}_{1}\right)=135 \mathrm{~Hz}$ $\mathrm{f}_{2}=165 \mathrm{~Hz}$ The different between two frequencies, $\mathrm{f}_{2}-\mathrm{f}_{1}=165-135=30 \mathrm{~Hz}$ According to question, $\mathrm{f}_{2}-\mathrm{f}_{1}=2 \times \text { fundamental frequency }$ $30=2 \times \text { fundamental frequency }$ $\therefore$ Fundamental frequency $=15 \mathrm{~Hz}$
173149
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will
1 increase by a factor of 2
2 decrease by a factor of 2
3 decrease by a factor of 4
4 remains unchanged
Explanation:
C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 . We know that, Intensity of sound $(\mathrm{I})=\mathrm{A}^{2} \mathrm{n}^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{2}$ $=\left(\frac{\mathrm{A}}{2 \mathrm{~A}}\right)^{2}\left(\frac{4 \mathrm{n}}{\mathrm{n}}\right)^{2}$ $\mathrm{I}_{2}=2 \mathrm{~A}$ $\mathrm{n}_{1}=4 \mathrm{n}$ $=\frac{1}{4} \times 16=4$ $\therefore \quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{4}$
AIPMT- 1989
WAVES
173150
Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of $40 \mathrm{~cm}$ is equivalent to phase difference of $\mathbf{1 . 6}$ $\pi$. The frequency of this wave is
1 $165 \mathrm{~Hz}$
2 $150 \mathrm{~Hz}$
3 $660 \mathrm{~Hz}$
4 $330 \mathrm{~Hz}$
Explanation:
C Given, Velocity of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ Path difference $(\Delta \mathrm{x})=40 \mathrm{~cm}=0.4 \mathrm{~m}$ Phase different $(\phi)=1.6 \pi$ We know that, $\frac{\Delta \mathrm{x}}{\Delta \phi}=\frac{\lambda}{2 \pi}$ Put the value in formula, $\frac{0.4}{1.6 \pi}=\frac{\lambda}{2 \pi}$ $\lambda=0.5 \mathrm{~m}$ The frequency of wave, $=\frac{\mathrm{V}}{\lambda}$ $=\frac{330 \mathrm{~m} / \mathrm{s}}{0.5}$ $=660 \mathrm{~Hz}$
AIPMT-1990
WAVES
173151
In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 \mathrm{~s}$. The frequency of the wave is
1 $1.47 \mathrm{~Hz}$
2 $0.36 \mathrm{~Hz}$
3 $0.73 \mathrm{~Hz}$
4 $2.94 \mathrm{~Hz}$
Explanation:
A Maximum displacement time (t) $=0.170 \mathrm{~s}$ For one vibration, Time period, $(\mathrm{T})=4 \mathrm{t}=4 \times 0.17$ $=0.68 \mathrm{sec}$ Frequency $(f)=\frac{1}{T}$ $=\frac{1}{0.68}=1.47 \mathrm{~Hz}$
AIPMT-1998
WAVES
173152
A pipe closed at one end and open at the other end resonates with a sound of frequency $135 \mathrm{~Hz}$ and also with $165 \mathrm{~Hz}$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is
1 $15 \mathrm{~Hz}$
2 $60 \mathrm{~Hz}$
3 $7.5 \mathrm{~Hz}$
4 $30 \mathrm{~Hz}$
Explanation:
A Given, The sound frequency $\left(\mathrm{f}_{1}\right)=135 \mathrm{~Hz}$ $\mathrm{f}_{2}=165 \mathrm{~Hz}$ The different between two frequencies, $\mathrm{f}_{2}-\mathrm{f}_{1}=165-135=30 \mathrm{~Hz}$ According to question, $\mathrm{f}_{2}-\mathrm{f}_{1}=2 \times \text { fundamental frequency }$ $30=2 \times \text { fundamental frequency }$ $\therefore$ Fundamental frequency $=15 \mathrm{~Hz}$
