355020
The length of two open organ pipes are \(l\) and \((l+\delta l)\) respectively. Neglecting end corrections the frequency of beats between them will be approximately (Here \(v\) is the speed of sound)
355021
When two sound waves of frequency differing by more than \(10\;Hz\) reach our ear simultaneously
1 Interference of sound does not take place
2 Beats are not produced
3 Beats are produced but we cannot identify them
4 The waves destructively interfere
Explanation:
Persistence of human ear is \(0.1\;s\) \(\therefore\) More than 10 beats can't be heared
PHXI15:WAVES
355022
A tuning fork of known frequency 256 \(Hz\) makes 5 beats per second with a vibrating string of a guitar. Beat frequency decreases to 2 beats per second when the tension in the guitar string is increased. The original frequency of the guitar string was
1 258 \(Hz\)
2 254 \(Hz\)
3 261 \(Hz\)
4 251 \(Hz\)
Explanation:
Let the frequencies of tuning fork and guitar string be \({v_{1}}\) and \({v_{2}}\) respectively.\(\begin{aligned}\therefore \quad v_{2} & =v_{1} \pm 5=256 {~Hz} \pm 5 \\& =261 {~Hz} \text { or } 251 {~Hz}\end{aligned}\) Increase in the tension of a guitar string increases its frequency so \({v_{2}=251 {~Hz}}\). So correct option is (4)
PHXI15:WAVES
355023
Beats are produced by frequencies \(f_{1}\) and \(f_{2}\). The duration of time between two successive maxima or minima is equal to
1 \(\dfrac{1}{f_{1}+f_{2}}\)
2 \(\dfrac{2}{f_{1}-f_{2}}\)
3 \(\dfrac{2}{f_{1}+f_{2}}\)
4 \(\dfrac{1}{f_{1}-f_{2}}\)
Explanation:
Time interval between two successive beats \(T=\dfrac{1}{\text { Beat frequency }}=\dfrac{1}{f_{1}-f_{2}}\)
PHXI15:WAVES
355024
Two waves \(Y_{1}=0.25 \sin 316 t\) and \(Y_{2}=0.25 \sin 310 t\) are propagating along the same direction. The number of beats produced per second are
1 \(\dfrac{\pi}{2}\)
2 \(\dfrac{2}{\pi}\)
3 \(\dfrac{3}{\pi}\)
4 \(\dfrac{\pi}{3}\)
Explanation:
Given, \({y_1} = 0.25\sin 316t\) and \(y_{2}=0.25\) \(\sin 310 t\) Comparing with the standard equation, \(y=A \sin \omega t\) we get \({\omega _1} = 316rad/s \Rightarrow 2\pi {f_1} = 316{\text{ or }}{f_1} = \frac{{316}}{{2\pi }}{H_Z}\) and \({\omega _2} = 310rad/s \Rightarrow 2\pi {f_2} = 310\,{\text{or}}\,{\text{ }}{f_2} = \frac{{310}}{{2\pi }}{H_Z}\) \(\therefore\) Beat frequency or number of beats produced per second \(f_{1}-f_{2}=\dfrac{316-310}{2 \pi}=\dfrac{6}{2 \pi}=\dfrac{3}{\pi}\)
355020
The length of two open organ pipes are \(l\) and \((l+\delta l)\) respectively. Neglecting end corrections the frequency of beats between them will be approximately (Here \(v\) is the speed of sound)
355021
When two sound waves of frequency differing by more than \(10\;Hz\) reach our ear simultaneously
1 Interference of sound does not take place
2 Beats are not produced
3 Beats are produced but we cannot identify them
4 The waves destructively interfere
Explanation:
Persistence of human ear is \(0.1\;s\) \(\therefore\) More than 10 beats can't be heared
PHXI15:WAVES
355022
A tuning fork of known frequency 256 \(Hz\) makes 5 beats per second with a vibrating string of a guitar. Beat frequency decreases to 2 beats per second when the tension in the guitar string is increased. The original frequency of the guitar string was
1 258 \(Hz\)
2 254 \(Hz\)
3 261 \(Hz\)
4 251 \(Hz\)
Explanation:
Let the frequencies of tuning fork and guitar string be \({v_{1}}\) and \({v_{2}}\) respectively.\(\begin{aligned}\therefore \quad v_{2} & =v_{1} \pm 5=256 {~Hz} \pm 5 \\& =261 {~Hz} \text { or } 251 {~Hz}\end{aligned}\) Increase in the tension of a guitar string increases its frequency so \({v_{2}=251 {~Hz}}\). So correct option is (4)
PHXI15:WAVES
355023
Beats are produced by frequencies \(f_{1}\) and \(f_{2}\). The duration of time between two successive maxima or minima is equal to
1 \(\dfrac{1}{f_{1}+f_{2}}\)
2 \(\dfrac{2}{f_{1}-f_{2}}\)
3 \(\dfrac{2}{f_{1}+f_{2}}\)
4 \(\dfrac{1}{f_{1}-f_{2}}\)
Explanation:
Time interval between two successive beats \(T=\dfrac{1}{\text { Beat frequency }}=\dfrac{1}{f_{1}-f_{2}}\)
PHXI15:WAVES
355024
Two waves \(Y_{1}=0.25 \sin 316 t\) and \(Y_{2}=0.25 \sin 310 t\) are propagating along the same direction. The number of beats produced per second are
1 \(\dfrac{\pi}{2}\)
2 \(\dfrac{2}{\pi}\)
3 \(\dfrac{3}{\pi}\)
4 \(\dfrac{\pi}{3}\)
Explanation:
Given, \({y_1} = 0.25\sin 316t\) and \(y_{2}=0.25\) \(\sin 310 t\) Comparing with the standard equation, \(y=A \sin \omega t\) we get \({\omega _1} = 316rad/s \Rightarrow 2\pi {f_1} = 316{\text{ or }}{f_1} = \frac{{316}}{{2\pi }}{H_Z}\) and \({\omega _2} = 310rad/s \Rightarrow 2\pi {f_2} = 310\,{\text{or}}\,{\text{ }}{f_2} = \frac{{310}}{{2\pi }}{H_Z}\) \(\therefore\) Beat frequency or number of beats produced per second \(f_{1}-f_{2}=\dfrac{316-310}{2 \pi}=\dfrac{6}{2 \pi}=\dfrac{3}{\pi}\)
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PHXI15:WAVES
355020
The length of two open organ pipes are \(l\) and \((l+\delta l)\) respectively. Neglecting end corrections the frequency of beats between them will be approximately (Here \(v\) is the speed of sound)
355021
When two sound waves of frequency differing by more than \(10\;Hz\) reach our ear simultaneously
1 Interference of sound does not take place
2 Beats are not produced
3 Beats are produced but we cannot identify them
4 The waves destructively interfere
Explanation:
Persistence of human ear is \(0.1\;s\) \(\therefore\) More than 10 beats can't be heared
PHXI15:WAVES
355022
A tuning fork of known frequency 256 \(Hz\) makes 5 beats per second with a vibrating string of a guitar. Beat frequency decreases to 2 beats per second when the tension in the guitar string is increased. The original frequency of the guitar string was
1 258 \(Hz\)
2 254 \(Hz\)
3 261 \(Hz\)
4 251 \(Hz\)
Explanation:
Let the frequencies of tuning fork and guitar string be \({v_{1}}\) and \({v_{2}}\) respectively.\(\begin{aligned}\therefore \quad v_{2} & =v_{1} \pm 5=256 {~Hz} \pm 5 \\& =261 {~Hz} \text { or } 251 {~Hz}\end{aligned}\) Increase in the tension of a guitar string increases its frequency so \({v_{2}=251 {~Hz}}\). So correct option is (4)
PHXI15:WAVES
355023
Beats are produced by frequencies \(f_{1}\) and \(f_{2}\). The duration of time between two successive maxima or minima is equal to
1 \(\dfrac{1}{f_{1}+f_{2}}\)
2 \(\dfrac{2}{f_{1}-f_{2}}\)
3 \(\dfrac{2}{f_{1}+f_{2}}\)
4 \(\dfrac{1}{f_{1}-f_{2}}\)
Explanation:
Time interval between two successive beats \(T=\dfrac{1}{\text { Beat frequency }}=\dfrac{1}{f_{1}-f_{2}}\)
PHXI15:WAVES
355024
Two waves \(Y_{1}=0.25 \sin 316 t\) and \(Y_{2}=0.25 \sin 310 t\) are propagating along the same direction. The number of beats produced per second are
1 \(\dfrac{\pi}{2}\)
2 \(\dfrac{2}{\pi}\)
3 \(\dfrac{3}{\pi}\)
4 \(\dfrac{\pi}{3}\)
Explanation:
Given, \({y_1} = 0.25\sin 316t\) and \(y_{2}=0.25\) \(\sin 310 t\) Comparing with the standard equation, \(y=A \sin \omega t\) we get \({\omega _1} = 316rad/s \Rightarrow 2\pi {f_1} = 316{\text{ or }}{f_1} = \frac{{316}}{{2\pi }}{H_Z}\) and \({\omega _2} = 310rad/s \Rightarrow 2\pi {f_2} = 310\,{\text{or}}\,{\text{ }}{f_2} = \frac{{310}}{{2\pi }}{H_Z}\) \(\therefore\) Beat frequency or number of beats produced per second \(f_{1}-f_{2}=\dfrac{316-310}{2 \pi}=\dfrac{6}{2 \pi}=\dfrac{3}{\pi}\)
355020
The length of two open organ pipes are \(l\) and \((l+\delta l)\) respectively. Neglecting end corrections the frequency of beats between them will be approximately (Here \(v\) is the speed of sound)
355021
When two sound waves of frequency differing by more than \(10\;Hz\) reach our ear simultaneously
1 Interference of sound does not take place
2 Beats are not produced
3 Beats are produced but we cannot identify them
4 The waves destructively interfere
Explanation:
Persistence of human ear is \(0.1\;s\) \(\therefore\) More than 10 beats can't be heared
PHXI15:WAVES
355022
A tuning fork of known frequency 256 \(Hz\) makes 5 beats per second with a vibrating string of a guitar. Beat frequency decreases to 2 beats per second when the tension in the guitar string is increased. The original frequency of the guitar string was
1 258 \(Hz\)
2 254 \(Hz\)
3 261 \(Hz\)
4 251 \(Hz\)
Explanation:
Let the frequencies of tuning fork and guitar string be \({v_{1}}\) and \({v_{2}}\) respectively.\(\begin{aligned}\therefore \quad v_{2} & =v_{1} \pm 5=256 {~Hz} \pm 5 \\& =261 {~Hz} \text { or } 251 {~Hz}\end{aligned}\) Increase in the tension of a guitar string increases its frequency so \({v_{2}=251 {~Hz}}\). So correct option is (4)
PHXI15:WAVES
355023
Beats are produced by frequencies \(f_{1}\) and \(f_{2}\). The duration of time between two successive maxima or minima is equal to
1 \(\dfrac{1}{f_{1}+f_{2}}\)
2 \(\dfrac{2}{f_{1}-f_{2}}\)
3 \(\dfrac{2}{f_{1}+f_{2}}\)
4 \(\dfrac{1}{f_{1}-f_{2}}\)
Explanation:
Time interval between two successive beats \(T=\dfrac{1}{\text { Beat frequency }}=\dfrac{1}{f_{1}-f_{2}}\)
PHXI15:WAVES
355024
Two waves \(Y_{1}=0.25 \sin 316 t\) and \(Y_{2}=0.25 \sin 310 t\) are propagating along the same direction. The number of beats produced per second are
1 \(\dfrac{\pi}{2}\)
2 \(\dfrac{2}{\pi}\)
3 \(\dfrac{3}{\pi}\)
4 \(\dfrac{\pi}{3}\)
Explanation:
Given, \({y_1} = 0.25\sin 316t\) and \(y_{2}=0.25\) \(\sin 310 t\) Comparing with the standard equation, \(y=A \sin \omega t\) we get \({\omega _1} = 316rad/s \Rightarrow 2\pi {f_1} = 316{\text{ or }}{f_1} = \frac{{316}}{{2\pi }}{H_Z}\) and \({\omega _2} = 310rad/s \Rightarrow 2\pi {f_2} = 310\,{\text{or}}\,{\text{ }}{f_2} = \frac{{310}}{{2\pi }}{H_Z}\) \(\therefore\) Beat frequency or number of beats produced per second \(f_{1}-f_{2}=\dfrac{316-310}{2 \pi}=\dfrac{6}{2 \pi}=\dfrac{3}{\pi}\)
355020
The length of two open organ pipes are \(l\) and \((l+\delta l)\) respectively. Neglecting end corrections the frequency of beats between them will be approximately (Here \(v\) is the speed of sound)
355021
When two sound waves of frequency differing by more than \(10\;Hz\) reach our ear simultaneously
1 Interference of sound does not take place
2 Beats are not produced
3 Beats are produced but we cannot identify them
4 The waves destructively interfere
Explanation:
Persistence of human ear is \(0.1\;s\) \(\therefore\) More than 10 beats can't be heared
PHXI15:WAVES
355022
A tuning fork of known frequency 256 \(Hz\) makes 5 beats per second with a vibrating string of a guitar. Beat frequency decreases to 2 beats per second when the tension in the guitar string is increased. The original frequency of the guitar string was
1 258 \(Hz\)
2 254 \(Hz\)
3 261 \(Hz\)
4 251 \(Hz\)
Explanation:
Let the frequencies of tuning fork and guitar string be \({v_{1}}\) and \({v_{2}}\) respectively.\(\begin{aligned}\therefore \quad v_{2} & =v_{1} \pm 5=256 {~Hz} \pm 5 \\& =261 {~Hz} \text { or } 251 {~Hz}\end{aligned}\) Increase in the tension of a guitar string increases its frequency so \({v_{2}=251 {~Hz}}\). So correct option is (4)
PHXI15:WAVES
355023
Beats are produced by frequencies \(f_{1}\) and \(f_{2}\). The duration of time between two successive maxima or minima is equal to
1 \(\dfrac{1}{f_{1}+f_{2}}\)
2 \(\dfrac{2}{f_{1}-f_{2}}\)
3 \(\dfrac{2}{f_{1}+f_{2}}\)
4 \(\dfrac{1}{f_{1}-f_{2}}\)
Explanation:
Time interval between two successive beats \(T=\dfrac{1}{\text { Beat frequency }}=\dfrac{1}{f_{1}-f_{2}}\)
PHXI15:WAVES
355024
Two waves \(Y_{1}=0.25 \sin 316 t\) and \(Y_{2}=0.25 \sin 310 t\) are propagating along the same direction. The number of beats produced per second are
1 \(\dfrac{\pi}{2}\)
2 \(\dfrac{2}{\pi}\)
3 \(\dfrac{3}{\pi}\)
4 \(\dfrac{\pi}{3}\)
Explanation:
Given, \({y_1} = 0.25\sin 316t\) and \(y_{2}=0.25\) \(\sin 310 t\) Comparing with the standard equation, \(y=A \sin \omega t\) we get \({\omega _1} = 316rad/s \Rightarrow 2\pi {f_1} = 316{\text{ or }}{f_1} = \frac{{316}}{{2\pi }}{H_Z}\) and \({\omega _2} = 310rad/s \Rightarrow 2\pi {f_2} = 310\,{\text{or}}\,{\text{ }}{f_2} = \frac{{310}}{{2\pi }}{H_Z}\) \(\therefore\) Beat frequency or number of beats produced per second \(f_{1}-f_{2}=\dfrac{316-310}{2 \pi}=\dfrac{6}{2 \pi}=\dfrac{3}{\pi}\)