The equation of vibration of a taut string, fixed at both ends, is given by; \(y=(3~\text{mm})~\text{cos}\left(\dfrac{\pi x}{10~\text{cm}}\right)~\text{sin}(800\pi~\text{s}^{-1}{t}).\) The positions of the nodes are:
1. \(x= 0~\text{cm}, 10~\text{cm}, 20~\text{cm},....\)
2. \(x= 0~\text{cm}, 20~\text{cm}, 40~\text{cm},....\)
3. \(x= 5~\text{cm}, 10~\text{cm}, 15~\text{cm},....\)
4. \(x= 5~\text{cm}, 15~\text{cm}, 25~\text{cm},....\)
Subtopic:  Beats |
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NEET 2026 - Target Batch - Vital
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Two different sources of sound having slightly different periods of vibration: \(1~\text{ms}\) and \(1.01~\text{ms},\) are sounded together. The resulting beat frequency is nearly:
1. \(100~\text{Hz}\)
2. \(50~\text{Hz}\)
3. \(10~\text{Hz}\)
4. \(0.01~\text{Hz}\)
Subtopic:  Beats |
From NCERT
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The correct figure that shows, schematically, the wave pattern produced by the superposition of two waves of frequencies \(9~\text{Hz}\) and \(11~\text{Hz}\), is:

1.
2.
3.
4.
Subtopic:  Beats |
 53%
From NCERT
JEE
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The addition of two waves with slightly different frequencies results in the production of beats or a beat frequency. The result can be monitored using an oscilloscope. Consider the two figures shown below:
                          
Keeping the lower frequency wave constant, we increase the frequency of the other wave. You would expect the display on the oscilloscope to go from:
1. Figure \(\mathrm I\) to Figure \(\mathrm{II}\).
2. Figure \(\mathrm{II}\) to Figure \(\mathrm{I}\).
3. There will not be a shift between Figure \(\mathrm{I}\) and Figure \(\mathrm{II}\), but the amplitude will increase.
4. There will not be a shift between Figure \(\mathrm{I}\) and Figure \(\mathrm{II}\), but the display will become brighter.
Subtopic:  Beats |
 51%
From NCERT
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