355139
If you set up the ninth harmonic on a string fixed at a both ends, its frequency compared to the seventh harmonic
1 Equal
2 Higher
3 Lower
4 None of these
Explanation:
Conceptual Question
PHXI15:WAVES
355140
A standing wave \(y=A \sin \left(\dfrac{20}{3} \pi x\right) \cos\) \((1000 \pi t)\) is maintained in a taut string where \(x\) and \(y\) are expressed in meters. The distance between the successive points oscillating with the amplitude \(A/2\) between two adjacent nodes is
1 \(10\;cm\)
2 \(25\;cm\)
3 \(25\;cm\)
4 \(20\;cm\)
Explanation:
From the given equation the amplitude is \({A_R} = A\sin \left( {\frac{{20\pi }}{3}x} \right){A_R} = \frac{A}{2} = A\sin \left( {\frac{{20\pi }}{3}x} \right)\) \( \Rightarrow \sin \left( {\frac{{20\pi }}{3}x} \right) = \frac{1}{2}\) \(\frac{{20\pi }}{3}{x_1} = \frac{\pi }{6} \Rightarrow {x_1} = \frac{1}{{40}}m = 2.5\;cm\) \(\frac{{20\pi }}{3}{x_2} = \frac{\pi }{2} + \frac{\pi }{3} \Rightarrow {x_2} = 12.5\;cm\) at \(x = 0\) it is a node The distance between points having amplitudes \(\frac{A}{2}{\text{is }}10\;cm\)
PHXI15:WAVES
355141
The fifth harmonic for vibrations of a stretched string is shown in figure. How many nodes are present here?
1 6
2 4
3 10
4 5
Explanation:
Conceptual Question
PHXI15:WAVES
355142
The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
355139
If you set up the ninth harmonic on a string fixed at a both ends, its frequency compared to the seventh harmonic
1 Equal
2 Higher
3 Lower
4 None of these
Explanation:
Conceptual Question
PHXI15:WAVES
355140
A standing wave \(y=A \sin \left(\dfrac{20}{3} \pi x\right) \cos\) \((1000 \pi t)\) is maintained in a taut string where \(x\) and \(y\) are expressed in meters. The distance between the successive points oscillating with the amplitude \(A/2\) between two adjacent nodes is
1 \(10\;cm\)
2 \(25\;cm\)
3 \(25\;cm\)
4 \(20\;cm\)
Explanation:
From the given equation the amplitude is \({A_R} = A\sin \left( {\frac{{20\pi }}{3}x} \right){A_R} = \frac{A}{2} = A\sin \left( {\frac{{20\pi }}{3}x} \right)\) \( \Rightarrow \sin \left( {\frac{{20\pi }}{3}x} \right) = \frac{1}{2}\) \(\frac{{20\pi }}{3}{x_1} = \frac{\pi }{6} \Rightarrow {x_1} = \frac{1}{{40}}m = 2.5\;cm\) \(\frac{{20\pi }}{3}{x_2} = \frac{\pi }{2} + \frac{\pi }{3} \Rightarrow {x_2} = 12.5\;cm\) at \(x = 0\) it is a node The distance between points having amplitudes \(\frac{A}{2}{\text{is }}10\;cm\)
PHXI15:WAVES
355141
The fifth harmonic for vibrations of a stretched string is shown in figure. How many nodes are present here?
1 6
2 4
3 10
4 5
Explanation:
Conceptual Question
PHXI15:WAVES
355142
The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
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PHXI15:WAVES
355139
If you set up the ninth harmonic on a string fixed at a both ends, its frequency compared to the seventh harmonic
1 Equal
2 Higher
3 Lower
4 None of these
Explanation:
Conceptual Question
PHXI15:WAVES
355140
A standing wave \(y=A \sin \left(\dfrac{20}{3} \pi x\right) \cos\) \((1000 \pi t)\) is maintained in a taut string where \(x\) and \(y\) are expressed in meters. The distance between the successive points oscillating with the amplitude \(A/2\) between two adjacent nodes is
1 \(10\;cm\)
2 \(25\;cm\)
3 \(25\;cm\)
4 \(20\;cm\)
Explanation:
From the given equation the amplitude is \({A_R} = A\sin \left( {\frac{{20\pi }}{3}x} \right){A_R} = \frac{A}{2} = A\sin \left( {\frac{{20\pi }}{3}x} \right)\) \( \Rightarrow \sin \left( {\frac{{20\pi }}{3}x} \right) = \frac{1}{2}\) \(\frac{{20\pi }}{3}{x_1} = \frac{\pi }{6} \Rightarrow {x_1} = \frac{1}{{40}}m = 2.5\;cm\) \(\frac{{20\pi }}{3}{x_2} = \frac{\pi }{2} + \frac{\pi }{3} \Rightarrow {x_2} = 12.5\;cm\) at \(x = 0\) it is a node The distance between points having amplitudes \(\frac{A}{2}{\text{is }}10\;cm\)
PHXI15:WAVES
355141
The fifth harmonic for vibrations of a stretched string is shown in figure. How many nodes are present here?
1 6
2 4
3 10
4 5
Explanation:
Conceptual Question
PHXI15:WAVES
355142
The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
355139
If you set up the ninth harmonic on a string fixed at a both ends, its frequency compared to the seventh harmonic
1 Equal
2 Higher
3 Lower
4 None of these
Explanation:
Conceptual Question
PHXI15:WAVES
355140
A standing wave \(y=A \sin \left(\dfrac{20}{3} \pi x\right) \cos\) \((1000 \pi t)\) is maintained in a taut string where \(x\) and \(y\) are expressed in meters. The distance between the successive points oscillating with the amplitude \(A/2\) between two adjacent nodes is
1 \(10\;cm\)
2 \(25\;cm\)
3 \(25\;cm\)
4 \(20\;cm\)
Explanation:
From the given equation the amplitude is \({A_R} = A\sin \left( {\frac{{20\pi }}{3}x} \right){A_R} = \frac{A}{2} = A\sin \left( {\frac{{20\pi }}{3}x} \right)\) \( \Rightarrow \sin \left( {\frac{{20\pi }}{3}x} \right) = \frac{1}{2}\) \(\frac{{20\pi }}{3}{x_1} = \frac{\pi }{6} \Rightarrow {x_1} = \frac{1}{{40}}m = 2.5\;cm\) \(\frac{{20\pi }}{3}{x_2} = \frac{\pi }{2} + \frac{\pi }{3} \Rightarrow {x_2} = 12.5\;cm\) at \(x = 0\) it is a node The distance between points having amplitudes \(\frac{A}{2}{\text{is }}10\;cm\)
PHXI15:WAVES
355141
The fifth harmonic for vibrations of a stretched string is shown in figure. How many nodes are present here?
1 6
2 4
3 10
4 5
Explanation:
Conceptual Question
PHXI15:WAVES
355142
The amplitude of vibration of any particle in a standing wave, produced along a stretched string depends on
