354990
A tuning fork of frequency \(512\;Hz\) makes 4 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
1 \(510\;Hz\)
2 \(514\;Hz\)
3 \(516\;Hz\)
4 \(508\;Hz\)
Explanation:
Suppose \(n_{p}=\) frequency of piano \(=\) ? \(\left(n_{p} \propto \sqrt{T}\right)\) \(n_{f}=\) frequency of tuning fork \( = 512\;Hz\) \(x=\) Beat frequency \(=4\) beats \(/ \mathrm{s}\), which is decreasing \((4 \rightarrow 2)\) after changing the tension of piano wire. Also, tension of piano wire is increasing so \(n_{p} \uparrow\) Hence, \({n_p} = 508\;Hz\)
PHXI15:WAVES
354991
For formation of beats, two sound notes must have
1 Different amplitudes and different frequencies
2 Exactly equal frequencies only
3 Exactly equal amplitudes only
4 Nearly equal frequencies and equal amplitudes
Explanation:
In sound notes various frequencies are mixed, hence beats can produce by two sound notes only when they have nearly equal frequencies and equal amplitudes
MHTCET - 2019
PHXI15:WAVES
354992
A tuning fork \(A\) produces 4 beats per second with another tuning fork \(B\) of frequency \(320\;Hz\). On filing one of the prongs of \(A\), 4 beats per second are again heard when sounded with the same fork \(B\). Then the frequency of the fork \(A\) before filing is
1 \(328\;Hz\)
2 \(316\;Hz\)
3 \(324\;Hz\)
4 \(320\;Hz\)
Explanation:
From the given condition of the problem, the frequencies of the two tuning forks \(A\) and \(B\) are related as \(n_{A}-n_{B}= \pm 4\) \( \Rightarrow {n_A} = {n_B} \pm 4 = 320 \pm 4 = 324\;Hz\) or 316 The frequency of \(A\) increases on filing. Hence, frequency of A before filing is \(316\;Hz.\)
PHXI15:WAVES
354993
A tuning fork of frequency \(200\;Hz\) is in unison with a sonometer wire. How many beats are heard in \(30\;s\) if the tension is increased by \(1 \%\)
1 10
2 20
3 30
4 5
Explanation:
\(\dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{\Delta T}{T} \Rightarrow \dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{1}{100}=\dfrac{1}{200}\) \(\Delta f=\dfrac{f}{200}=1\) \(\therefore\) no.of beats \(=1\) So 30 beats are heard in \(30\;s\).
PHXI15:WAVES
354994
Wire having tension \(225\,N\) produces six beats per second when it is turned with a fork. When tension changes to \(256\,N\), it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be
1 \(186\;Hz\)
2 \(225\;Hz\)
3 \(256\;Hz\)
4 \(280\;Hz\)
Explanation:
The fundamental frequency of wire vibrating under tension \(T\) is given by \(f_{1}=\dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}\) Here, \(\mu=\) mass of string per unit length Let the frequency of tunning fork be \(x\) which beat 6 beats per second. Let \((x-f)= \pm 6\) \(\therefore f_{1}=\dfrac{1}{2 L} \sqrt{\dfrac{225}{\mu}}\) and \(f_{2}=\dfrac{1}{2 L} \sqrt{\dfrac{256}{\mu}}\) \(\therefore \dfrac{f_{1}}{f_{2}}=\dfrac{15}{16}\) \(\therefore f_{2}=\dfrac{16}{15} \times f_{1}\) \(\Rightarrow f_{2}=\dfrac{16}{15}(6+x)\) taking two cases of \(f_{1}\) and equating both, we have \((x+6)=\dfrac{16}{15}(x-6)\) \(\therefore 15 x+90=16 x-96\) \(\therefore x = 186\;Hz\)
354990
A tuning fork of frequency \(512\;Hz\) makes 4 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
1 \(510\;Hz\)
2 \(514\;Hz\)
3 \(516\;Hz\)
4 \(508\;Hz\)
Explanation:
Suppose \(n_{p}=\) frequency of piano \(=\) ? \(\left(n_{p} \propto \sqrt{T}\right)\) \(n_{f}=\) frequency of tuning fork \( = 512\;Hz\) \(x=\) Beat frequency \(=4\) beats \(/ \mathrm{s}\), which is decreasing \((4 \rightarrow 2)\) after changing the tension of piano wire. Also, tension of piano wire is increasing so \(n_{p} \uparrow\) Hence, \({n_p} = 508\;Hz\)
PHXI15:WAVES
354991
For formation of beats, two sound notes must have
1 Different amplitudes and different frequencies
2 Exactly equal frequencies only
3 Exactly equal amplitudes only
4 Nearly equal frequencies and equal amplitudes
Explanation:
In sound notes various frequencies are mixed, hence beats can produce by two sound notes only when they have nearly equal frequencies and equal amplitudes
MHTCET - 2019
PHXI15:WAVES
354992
A tuning fork \(A\) produces 4 beats per second with another tuning fork \(B\) of frequency \(320\;Hz\). On filing one of the prongs of \(A\), 4 beats per second are again heard when sounded with the same fork \(B\). Then the frequency of the fork \(A\) before filing is
1 \(328\;Hz\)
2 \(316\;Hz\)
3 \(324\;Hz\)
4 \(320\;Hz\)
Explanation:
From the given condition of the problem, the frequencies of the two tuning forks \(A\) and \(B\) are related as \(n_{A}-n_{B}= \pm 4\) \( \Rightarrow {n_A} = {n_B} \pm 4 = 320 \pm 4 = 324\;Hz\) or 316 The frequency of \(A\) increases on filing. Hence, frequency of A before filing is \(316\;Hz.\)
PHXI15:WAVES
354993
A tuning fork of frequency \(200\;Hz\) is in unison with a sonometer wire. How many beats are heard in \(30\;s\) if the tension is increased by \(1 \%\)
1 10
2 20
3 30
4 5
Explanation:
\(\dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{\Delta T}{T} \Rightarrow \dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{1}{100}=\dfrac{1}{200}\) \(\Delta f=\dfrac{f}{200}=1\) \(\therefore\) no.of beats \(=1\) So 30 beats are heard in \(30\;s\).
PHXI15:WAVES
354994
Wire having tension \(225\,N\) produces six beats per second when it is turned with a fork. When tension changes to \(256\,N\), it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be
1 \(186\;Hz\)
2 \(225\;Hz\)
3 \(256\;Hz\)
4 \(280\;Hz\)
Explanation:
The fundamental frequency of wire vibrating under tension \(T\) is given by \(f_{1}=\dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}\) Here, \(\mu=\) mass of string per unit length Let the frequency of tunning fork be \(x\) which beat 6 beats per second. Let \((x-f)= \pm 6\) \(\therefore f_{1}=\dfrac{1}{2 L} \sqrt{\dfrac{225}{\mu}}\) and \(f_{2}=\dfrac{1}{2 L} \sqrt{\dfrac{256}{\mu}}\) \(\therefore \dfrac{f_{1}}{f_{2}}=\dfrac{15}{16}\) \(\therefore f_{2}=\dfrac{16}{15} \times f_{1}\) \(\Rightarrow f_{2}=\dfrac{16}{15}(6+x)\) taking two cases of \(f_{1}\) and equating both, we have \((x+6)=\dfrac{16}{15}(x-6)\) \(\therefore 15 x+90=16 x-96\) \(\therefore x = 186\;Hz\)
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PHXI15:WAVES
354990
A tuning fork of frequency \(512\;Hz\) makes 4 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
1 \(510\;Hz\)
2 \(514\;Hz\)
3 \(516\;Hz\)
4 \(508\;Hz\)
Explanation:
Suppose \(n_{p}=\) frequency of piano \(=\) ? \(\left(n_{p} \propto \sqrt{T}\right)\) \(n_{f}=\) frequency of tuning fork \( = 512\;Hz\) \(x=\) Beat frequency \(=4\) beats \(/ \mathrm{s}\), which is decreasing \((4 \rightarrow 2)\) after changing the tension of piano wire. Also, tension of piano wire is increasing so \(n_{p} \uparrow\) Hence, \({n_p} = 508\;Hz\)
PHXI15:WAVES
354991
For formation of beats, two sound notes must have
1 Different amplitudes and different frequencies
2 Exactly equal frequencies only
3 Exactly equal amplitudes only
4 Nearly equal frequencies and equal amplitudes
Explanation:
In sound notes various frequencies are mixed, hence beats can produce by two sound notes only when they have nearly equal frequencies and equal amplitudes
MHTCET - 2019
PHXI15:WAVES
354992
A tuning fork \(A\) produces 4 beats per second with another tuning fork \(B\) of frequency \(320\;Hz\). On filing one of the prongs of \(A\), 4 beats per second are again heard when sounded with the same fork \(B\). Then the frequency of the fork \(A\) before filing is
1 \(328\;Hz\)
2 \(316\;Hz\)
3 \(324\;Hz\)
4 \(320\;Hz\)
Explanation:
From the given condition of the problem, the frequencies of the two tuning forks \(A\) and \(B\) are related as \(n_{A}-n_{B}= \pm 4\) \( \Rightarrow {n_A} = {n_B} \pm 4 = 320 \pm 4 = 324\;Hz\) or 316 The frequency of \(A\) increases on filing. Hence, frequency of A before filing is \(316\;Hz.\)
PHXI15:WAVES
354993
A tuning fork of frequency \(200\;Hz\) is in unison with a sonometer wire. How many beats are heard in \(30\;s\) if the tension is increased by \(1 \%\)
1 10
2 20
3 30
4 5
Explanation:
\(\dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{\Delta T}{T} \Rightarrow \dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{1}{100}=\dfrac{1}{200}\) \(\Delta f=\dfrac{f}{200}=1\) \(\therefore\) no.of beats \(=1\) So 30 beats are heard in \(30\;s\).
PHXI15:WAVES
354994
Wire having tension \(225\,N\) produces six beats per second when it is turned with a fork. When tension changes to \(256\,N\), it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be
1 \(186\;Hz\)
2 \(225\;Hz\)
3 \(256\;Hz\)
4 \(280\;Hz\)
Explanation:
The fundamental frequency of wire vibrating under tension \(T\) is given by \(f_{1}=\dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}\) Here, \(\mu=\) mass of string per unit length Let the frequency of tunning fork be \(x\) which beat 6 beats per second. Let \((x-f)= \pm 6\) \(\therefore f_{1}=\dfrac{1}{2 L} \sqrt{\dfrac{225}{\mu}}\) and \(f_{2}=\dfrac{1}{2 L} \sqrt{\dfrac{256}{\mu}}\) \(\therefore \dfrac{f_{1}}{f_{2}}=\dfrac{15}{16}\) \(\therefore f_{2}=\dfrac{16}{15} \times f_{1}\) \(\Rightarrow f_{2}=\dfrac{16}{15}(6+x)\) taking two cases of \(f_{1}\) and equating both, we have \((x+6)=\dfrac{16}{15}(x-6)\) \(\therefore 15 x+90=16 x-96\) \(\therefore x = 186\;Hz\)
354990
A tuning fork of frequency \(512\;Hz\) makes 4 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
1 \(510\;Hz\)
2 \(514\;Hz\)
3 \(516\;Hz\)
4 \(508\;Hz\)
Explanation:
Suppose \(n_{p}=\) frequency of piano \(=\) ? \(\left(n_{p} \propto \sqrt{T}\right)\) \(n_{f}=\) frequency of tuning fork \( = 512\;Hz\) \(x=\) Beat frequency \(=4\) beats \(/ \mathrm{s}\), which is decreasing \((4 \rightarrow 2)\) after changing the tension of piano wire. Also, tension of piano wire is increasing so \(n_{p} \uparrow\) Hence, \({n_p} = 508\;Hz\)
PHXI15:WAVES
354991
For formation of beats, two sound notes must have
1 Different amplitudes and different frequencies
2 Exactly equal frequencies only
3 Exactly equal amplitudes only
4 Nearly equal frequencies and equal amplitudes
Explanation:
In sound notes various frequencies are mixed, hence beats can produce by two sound notes only when they have nearly equal frequencies and equal amplitudes
MHTCET - 2019
PHXI15:WAVES
354992
A tuning fork \(A\) produces 4 beats per second with another tuning fork \(B\) of frequency \(320\;Hz\). On filing one of the prongs of \(A\), 4 beats per second are again heard when sounded with the same fork \(B\). Then the frequency of the fork \(A\) before filing is
1 \(328\;Hz\)
2 \(316\;Hz\)
3 \(324\;Hz\)
4 \(320\;Hz\)
Explanation:
From the given condition of the problem, the frequencies of the two tuning forks \(A\) and \(B\) are related as \(n_{A}-n_{B}= \pm 4\) \( \Rightarrow {n_A} = {n_B} \pm 4 = 320 \pm 4 = 324\;Hz\) or 316 The frequency of \(A\) increases on filing. Hence, frequency of A before filing is \(316\;Hz.\)
PHXI15:WAVES
354993
A tuning fork of frequency \(200\;Hz\) is in unison with a sonometer wire. How many beats are heard in \(30\;s\) if the tension is increased by \(1 \%\)
1 10
2 20
3 30
4 5
Explanation:
\(\dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{\Delta T}{T} \Rightarrow \dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{1}{100}=\dfrac{1}{200}\) \(\Delta f=\dfrac{f}{200}=1\) \(\therefore\) no.of beats \(=1\) So 30 beats are heard in \(30\;s\).
PHXI15:WAVES
354994
Wire having tension \(225\,N\) produces six beats per second when it is turned with a fork. When tension changes to \(256\,N\), it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be
1 \(186\;Hz\)
2 \(225\;Hz\)
3 \(256\;Hz\)
4 \(280\;Hz\)
Explanation:
The fundamental frequency of wire vibrating under tension \(T\) is given by \(f_{1}=\dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}\) Here, \(\mu=\) mass of string per unit length Let the frequency of tunning fork be \(x\) which beat 6 beats per second. Let \((x-f)= \pm 6\) \(\therefore f_{1}=\dfrac{1}{2 L} \sqrt{\dfrac{225}{\mu}}\) and \(f_{2}=\dfrac{1}{2 L} \sqrt{\dfrac{256}{\mu}}\) \(\therefore \dfrac{f_{1}}{f_{2}}=\dfrac{15}{16}\) \(\therefore f_{2}=\dfrac{16}{15} \times f_{1}\) \(\Rightarrow f_{2}=\dfrac{16}{15}(6+x)\) taking two cases of \(f_{1}\) and equating both, we have \((x+6)=\dfrac{16}{15}(x-6)\) \(\therefore 15 x+90=16 x-96\) \(\therefore x = 186\;Hz\)
354990
A tuning fork of frequency \(512\;Hz\) makes 4 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
1 \(510\;Hz\)
2 \(514\;Hz\)
3 \(516\;Hz\)
4 \(508\;Hz\)
Explanation:
Suppose \(n_{p}=\) frequency of piano \(=\) ? \(\left(n_{p} \propto \sqrt{T}\right)\) \(n_{f}=\) frequency of tuning fork \( = 512\;Hz\) \(x=\) Beat frequency \(=4\) beats \(/ \mathrm{s}\), which is decreasing \((4 \rightarrow 2)\) after changing the tension of piano wire. Also, tension of piano wire is increasing so \(n_{p} \uparrow\) Hence, \({n_p} = 508\;Hz\)
PHXI15:WAVES
354991
For formation of beats, two sound notes must have
1 Different amplitudes and different frequencies
2 Exactly equal frequencies only
3 Exactly equal amplitudes only
4 Nearly equal frequencies and equal amplitudes
Explanation:
In sound notes various frequencies are mixed, hence beats can produce by two sound notes only when they have nearly equal frequencies and equal amplitudes
MHTCET - 2019
PHXI15:WAVES
354992
A tuning fork \(A\) produces 4 beats per second with another tuning fork \(B\) of frequency \(320\;Hz\). On filing one of the prongs of \(A\), 4 beats per second are again heard when sounded with the same fork \(B\). Then the frequency of the fork \(A\) before filing is
1 \(328\;Hz\)
2 \(316\;Hz\)
3 \(324\;Hz\)
4 \(320\;Hz\)
Explanation:
From the given condition of the problem, the frequencies of the two tuning forks \(A\) and \(B\) are related as \(n_{A}-n_{B}= \pm 4\) \( \Rightarrow {n_A} = {n_B} \pm 4 = 320 \pm 4 = 324\;Hz\) or 316 The frequency of \(A\) increases on filing. Hence, frequency of A before filing is \(316\;Hz.\)
PHXI15:WAVES
354993
A tuning fork of frequency \(200\;Hz\) is in unison with a sonometer wire. How many beats are heard in \(30\;s\) if the tension is increased by \(1 \%\)
1 10
2 20
3 30
4 5
Explanation:
\(\dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{\Delta T}{T} \Rightarrow \dfrac{\Delta f}{f}=\dfrac{1}{2} \dfrac{1}{100}=\dfrac{1}{200}\) \(\Delta f=\dfrac{f}{200}=1\) \(\therefore\) no.of beats \(=1\) So 30 beats are heard in \(30\;s\).
PHXI15:WAVES
354994
Wire having tension \(225\,N\) produces six beats per second when it is turned with a fork. When tension changes to \(256\,N\), it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be
1 \(186\;Hz\)
2 \(225\;Hz\)
3 \(256\;Hz\)
4 \(280\;Hz\)
Explanation:
The fundamental frequency of wire vibrating under tension \(T\) is given by \(f_{1}=\dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}\) Here, \(\mu=\) mass of string per unit length Let the frequency of tunning fork be \(x\) which beat 6 beats per second. Let \((x-f)= \pm 6\) \(\therefore f_{1}=\dfrac{1}{2 L} \sqrt{\dfrac{225}{\mu}}\) and \(f_{2}=\dfrac{1}{2 L} \sqrt{\dfrac{256}{\mu}}\) \(\therefore \dfrac{f_{1}}{f_{2}}=\dfrac{15}{16}\) \(\therefore f_{2}=\dfrac{16}{15} \times f_{1}\) \(\Rightarrow f_{2}=\dfrac{16}{15}(6+x)\) taking two cases of \(f_{1}\) and equating both, we have \((x+6)=\dfrac{16}{15}(x-6)\) \(\therefore 15 x+90=16 x-96\) \(\therefore x = 186\;Hz\)