355177
A string is stretched between two fixed points separated by \(75\,cm\). It is observed to have resonant frequencies at \(420\,Hz\) and \(315\,Hz\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is
1 \(10.5\,Hz\)
2 \(105\,Hz\)
3 \(1.05\,Hz\)
4 \(1050\,Hz\)
Explanation:
\({\left(\dfrac{n}{2 l}\right) v=315}\) and \({\dfrac{(n+1) v}{2 l}=420}\) \({\Rightarrow n=3}\) Lowest frequency \({f_{0}=\dfrac{v}{2 l}=\dfrac{315}{3}=105 {~Hz}}\)
PHXI15:WAVES
355178
A string with a mass density of \(4 \times {10^{ - 3}}\;kg/m\) is under tension of \(360\;N\) and is fixed at both ends. One of its resonance frequencies is \(375\;Hz\). The next higher resonance frequency is\(450\;Hz\). The mass of the string is :-
355179
When stationary waves are set up, pick out the correct statement from the following
1 All the particles in the medium are in the same phase of vibration at all times and distances
2 The particles between two consecutive nodes are in phase.
3 The phase lag along the path of the wave increases as the distance from the source increases
4 Only antinodes are in same phase
Explanation:
Conceptual Question
PHXI15:WAVES
355180
The first overtone of a stretched wire of given length in \(320\;Hz\). The first harmonic is
1 \(160\;Hz\)
2 \(320\;Hz\)
3 \(640\;Hz\)
4 \(480\;Hz\)
Explanation:
Frequency of first overtone are second harmonic \(\left( {{n_2}} \right) = 320\;Hz\). So, frequency of first harmonic \({n_1} = \frac{{{n_2}}}{2} = \frac{{320}}{2} = 160\;Hz\)
PHXI15:WAVES
355181
A travelling wave represented by \(y=A \sin (\omega t-k x)\) is superimposed on another wave represented by \(y=A \sin (\omega t+k x)\). The resultant is
1 A wave travelling along \(+x\) dircetion
2 A wave travelling along \(-x\) dircetion
3 A standing wave having nodes at \(x=\dfrac{n \lambda}{2}, n=0,1,2 \ldots\)
4 A standing wave having nodes at \(x=\left(n+\dfrac{1}{2}\right) \dfrac{\lambda}{2} ; n=0,1,2 \ldots\)
Explanation:
\(y=A \sin (\omega t-k x)+A \sin (\omega t+k x)\) \(y=2 A \sin \omega t \cos k x\) The amplitude is \(A_{R}=2 A \cos k x\) For standing wave nodes \(\begin{aligned}& \cos k x=0 \\& \dfrac{2 \pi}{\lambda} \cdot x=(2 n+1) \dfrac{\pi}{2} \\& \therefore x=\dfrac{(2 n+1) \lambda}{4}, n=0,1,2,3, \ldots . .\end{aligned}\)
355177
A string is stretched between two fixed points separated by \(75\,cm\). It is observed to have resonant frequencies at \(420\,Hz\) and \(315\,Hz\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is
1 \(10.5\,Hz\)
2 \(105\,Hz\)
3 \(1.05\,Hz\)
4 \(1050\,Hz\)
Explanation:
\({\left(\dfrac{n}{2 l}\right) v=315}\) and \({\dfrac{(n+1) v}{2 l}=420}\) \({\Rightarrow n=3}\) Lowest frequency \({f_{0}=\dfrac{v}{2 l}=\dfrac{315}{3}=105 {~Hz}}\)
PHXI15:WAVES
355178
A string with a mass density of \(4 \times {10^{ - 3}}\;kg/m\) is under tension of \(360\;N\) and is fixed at both ends. One of its resonance frequencies is \(375\;Hz\). The next higher resonance frequency is\(450\;Hz\). The mass of the string is :-
355179
When stationary waves are set up, pick out the correct statement from the following
1 All the particles in the medium are in the same phase of vibration at all times and distances
2 The particles between two consecutive nodes are in phase.
3 The phase lag along the path of the wave increases as the distance from the source increases
4 Only antinodes are in same phase
Explanation:
Conceptual Question
PHXI15:WAVES
355180
The first overtone of a stretched wire of given length in \(320\;Hz\). The first harmonic is
1 \(160\;Hz\)
2 \(320\;Hz\)
3 \(640\;Hz\)
4 \(480\;Hz\)
Explanation:
Frequency of first overtone are second harmonic \(\left( {{n_2}} \right) = 320\;Hz\). So, frequency of first harmonic \({n_1} = \frac{{{n_2}}}{2} = \frac{{320}}{2} = 160\;Hz\)
PHXI15:WAVES
355181
A travelling wave represented by \(y=A \sin (\omega t-k x)\) is superimposed on another wave represented by \(y=A \sin (\omega t+k x)\). The resultant is
1 A wave travelling along \(+x\) dircetion
2 A wave travelling along \(-x\) dircetion
3 A standing wave having nodes at \(x=\dfrac{n \lambda}{2}, n=0,1,2 \ldots\)
4 A standing wave having nodes at \(x=\left(n+\dfrac{1}{2}\right) \dfrac{\lambda}{2} ; n=0,1,2 \ldots\)
Explanation:
\(y=A \sin (\omega t-k x)+A \sin (\omega t+k x)\) \(y=2 A \sin \omega t \cos k x\) The amplitude is \(A_{R}=2 A \cos k x\) For standing wave nodes \(\begin{aligned}& \cos k x=0 \\& \dfrac{2 \pi}{\lambda} \cdot x=(2 n+1) \dfrac{\pi}{2} \\& \therefore x=\dfrac{(2 n+1) \lambda}{4}, n=0,1,2,3, \ldots . .\end{aligned}\)
355177
A string is stretched between two fixed points separated by \(75\,cm\). It is observed to have resonant frequencies at \(420\,Hz\) and \(315\,Hz\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is
1 \(10.5\,Hz\)
2 \(105\,Hz\)
3 \(1.05\,Hz\)
4 \(1050\,Hz\)
Explanation:
\({\left(\dfrac{n}{2 l}\right) v=315}\) and \({\dfrac{(n+1) v}{2 l}=420}\) \({\Rightarrow n=3}\) Lowest frequency \({f_{0}=\dfrac{v}{2 l}=\dfrac{315}{3}=105 {~Hz}}\)
PHXI15:WAVES
355178
A string with a mass density of \(4 \times {10^{ - 3}}\;kg/m\) is under tension of \(360\;N\) and is fixed at both ends. One of its resonance frequencies is \(375\;Hz\). The next higher resonance frequency is\(450\;Hz\). The mass of the string is :-
355179
When stationary waves are set up, pick out the correct statement from the following
1 All the particles in the medium are in the same phase of vibration at all times and distances
2 The particles between two consecutive nodes are in phase.
3 The phase lag along the path of the wave increases as the distance from the source increases
4 Only antinodes are in same phase
Explanation:
Conceptual Question
PHXI15:WAVES
355180
The first overtone of a stretched wire of given length in \(320\;Hz\). The first harmonic is
1 \(160\;Hz\)
2 \(320\;Hz\)
3 \(640\;Hz\)
4 \(480\;Hz\)
Explanation:
Frequency of first overtone are second harmonic \(\left( {{n_2}} \right) = 320\;Hz\). So, frequency of first harmonic \({n_1} = \frac{{{n_2}}}{2} = \frac{{320}}{2} = 160\;Hz\)
PHXI15:WAVES
355181
A travelling wave represented by \(y=A \sin (\omega t-k x)\) is superimposed on another wave represented by \(y=A \sin (\omega t+k x)\). The resultant is
1 A wave travelling along \(+x\) dircetion
2 A wave travelling along \(-x\) dircetion
3 A standing wave having nodes at \(x=\dfrac{n \lambda}{2}, n=0,1,2 \ldots\)
4 A standing wave having nodes at \(x=\left(n+\dfrac{1}{2}\right) \dfrac{\lambda}{2} ; n=0,1,2 \ldots\)
Explanation:
\(y=A \sin (\omega t-k x)+A \sin (\omega t+k x)\) \(y=2 A \sin \omega t \cos k x\) The amplitude is \(A_{R}=2 A \cos k x\) For standing wave nodes \(\begin{aligned}& \cos k x=0 \\& \dfrac{2 \pi}{\lambda} \cdot x=(2 n+1) \dfrac{\pi}{2} \\& \therefore x=\dfrac{(2 n+1) \lambda}{4}, n=0,1,2,3, \ldots . .\end{aligned}\)
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PHXI15:WAVES
355177
A string is stretched between two fixed points separated by \(75\,cm\). It is observed to have resonant frequencies at \(420\,Hz\) and \(315\,Hz\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is
1 \(10.5\,Hz\)
2 \(105\,Hz\)
3 \(1.05\,Hz\)
4 \(1050\,Hz\)
Explanation:
\({\left(\dfrac{n}{2 l}\right) v=315}\) and \({\dfrac{(n+1) v}{2 l}=420}\) \({\Rightarrow n=3}\) Lowest frequency \({f_{0}=\dfrac{v}{2 l}=\dfrac{315}{3}=105 {~Hz}}\)
PHXI15:WAVES
355178
A string with a mass density of \(4 \times {10^{ - 3}}\;kg/m\) is under tension of \(360\;N\) and is fixed at both ends. One of its resonance frequencies is \(375\;Hz\). The next higher resonance frequency is\(450\;Hz\). The mass of the string is :-
355179
When stationary waves are set up, pick out the correct statement from the following
1 All the particles in the medium are in the same phase of vibration at all times and distances
2 The particles between two consecutive nodes are in phase.
3 The phase lag along the path of the wave increases as the distance from the source increases
4 Only antinodes are in same phase
Explanation:
Conceptual Question
PHXI15:WAVES
355180
The first overtone of a stretched wire of given length in \(320\;Hz\). The first harmonic is
1 \(160\;Hz\)
2 \(320\;Hz\)
3 \(640\;Hz\)
4 \(480\;Hz\)
Explanation:
Frequency of first overtone are second harmonic \(\left( {{n_2}} \right) = 320\;Hz\). So, frequency of first harmonic \({n_1} = \frac{{{n_2}}}{2} = \frac{{320}}{2} = 160\;Hz\)
PHXI15:WAVES
355181
A travelling wave represented by \(y=A \sin (\omega t-k x)\) is superimposed on another wave represented by \(y=A \sin (\omega t+k x)\). The resultant is
1 A wave travelling along \(+x\) dircetion
2 A wave travelling along \(-x\) dircetion
3 A standing wave having nodes at \(x=\dfrac{n \lambda}{2}, n=0,1,2 \ldots\)
4 A standing wave having nodes at \(x=\left(n+\dfrac{1}{2}\right) \dfrac{\lambda}{2} ; n=0,1,2 \ldots\)
Explanation:
\(y=A \sin (\omega t-k x)+A \sin (\omega t+k x)\) \(y=2 A \sin \omega t \cos k x\) The amplitude is \(A_{R}=2 A \cos k x\) For standing wave nodes \(\begin{aligned}& \cos k x=0 \\& \dfrac{2 \pi}{\lambda} \cdot x=(2 n+1) \dfrac{\pi}{2} \\& \therefore x=\dfrac{(2 n+1) \lambda}{4}, n=0,1,2,3, \ldots . .\end{aligned}\)
355177
A string is stretched between two fixed points separated by \(75\,cm\). It is observed to have resonant frequencies at \(420\,Hz\) and \(315\,Hz\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is
1 \(10.5\,Hz\)
2 \(105\,Hz\)
3 \(1.05\,Hz\)
4 \(1050\,Hz\)
Explanation:
\({\left(\dfrac{n}{2 l}\right) v=315}\) and \({\dfrac{(n+1) v}{2 l}=420}\) \({\Rightarrow n=3}\) Lowest frequency \({f_{0}=\dfrac{v}{2 l}=\dfrac{315}{3}=105 {~Hz}}\)
PHXI15:WAVES
355178
A string with a mass density of \(4 \times {10^{ - 3}}\;kg/m\) is under tension of \(360\;N\) and is fixed at both ends. One of its resonance frequencies is \(375\;Hz\). The next higher resonance frequency is\(450\;Hz\). The mass of the string is :-
355179
When stationary waves are set up, pick out the correct statement from the following
1 All the particles in the medium are in the same phase of vibration at all times and distances
2 The particles between two consecutive nodes are in phase.
3 The phase lag along the path of the wave increases as the distance from the source increases
4 Only antinodes are in same phase
Explanation:
Conceptual Question
PHXI15:WAVES
355180
The first overtone of a stretched wire of given length in \(320\;Hz\). The first harmonic is
1 \(160\;Hz\)
2 \(320\;Hz\)
3 \(640\;Hz\)
4 \(480\;Hz\)
Explanation:
Frequency of first overtone are second harmonic \(\left( {{n_2}} \right) = 320\;Hz\). So, frequency of first harmonic \({n_1} = \frac{{{n_2}}}{2} = \frac{{320}}{2} = 160\;Hz\)
PHXI15:WAVES
355181
A travelling wave represented by \(y=A \sin (\omega t-k x)\) is superimposed on another wave represented by \(y=A \sin (\omega t+k x)\). The resultant is
1 A wave travelling along \(+x\) dircetion
2 A wave travelling along \(-x\) dircetion
3 A standing wave having nodes at \(x=\dfrac{n \lambda}{2}, n=0,1,2 \ldots\)
4 A standing wave having nodes at \(x=\left(n+\dfrac{1}{2}\right) \dfrac{\lambda}{2} ; n=0,1,2 \ldots\)
Explanation:
\(y=A \sin (\omega t-k x)+A \sin (\omega t+k x)\) \(y=2 A \sin \omega t \cos k x\) The amplitude is \(A_{R}=2 A \cos k x\) For standing wave nodes \(\begin{aligned}& \cos k x=0 \\& \dfrac{2 \pi}{\lambda} \cdot x=(2 n+1) \dfrac{\pi}{2} \\& \therefore x=\dfrac{(2 n+1) \lambda}{4}, n=0,1,2,3, \ldots . .\end{aligned}\)