354515
At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is
1 Sinusoidal shape with amplitude \(A\)
2 Straight line
3 Sinusoidal shape with amplitude \(A/3\)
4 Sinusoidal shape with amplitude \(A/2\)
Explanation:
Particles have kinetic energy maximum at mean position. So the string will be at mean position present as a straight line.
PHXI15:WAVES
354516
The equation \(y=4+2 \sin (6 t-3 x)\) represents a wave motion, then wave speed and amplitude, respectively are
1 Wave speed 1 unit, amplitude 6 units
2 Wave speed 2 units, amplitude 2 units
3 Wave speed 4 units, amplitude \(1 / 2\) unit
4 Wave speed \(1 / 2\) unit, amplitude 5 units
Explanation:
The shape of the wave is shown in the figure From this graph, we see that amplitude of wave is 2 units. Wave speed, \(v = \frac{{{\rm{Coefficient}}\,\,\,{\rm{of}}\,\,t}}{{{\rm{Coefficeint}}\,\,{\rm{of}}\,\,x}} = \frac{6}{3} = 2{\mkern 1mu} units\)
AIIMS - 2009
PHXI15:WAVES
354517
A wave of frequency \(500\;Hz\) has a velocity \(360\;m{s^{ - 1}}\). The phase difference between two displacements at a certain point at time \({10^{ - 3}}\;s\) apart will be
354518
The phase difference between two waves represented by \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(\quad y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) where \(x\) is expressed in metre and \(t\) is expressed in second, is approximately
1 \(1.07\,rad\)
2 \(2.07\,rad\)
3 \(0.5\,rad\)
4 \(1.5\,rad\)
Explanation:
The given waves equations are \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) \(\begin{array}{r}\Rightarrow y_{2}=10^{-6} \sin \left[100 t+(x / 50)+\dfrac{\pi}{2}\right] m \\{\left[\because \sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right]}\end{array}\) Hence, the phase difference between the waves is \(\Delta \phi = \left( {\frac{\pi }{2} - 0.5} \right)rad = \left( {\frac{{3.14}}{2} - 0.5} \right)rad\) \( = (1.57 - 0.5)\,rad = 1.07\,rad\)
354515
At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is
1 Sinusoidal shape with amplitude \(A\)
2 Straight line
3 Sinusoidal shape with amplitude \(A/3\)
4 Sinusoidal shape with amplitude \(A/2\)
Explanation:
Particles have kinetic energy maximum at mean position. So the string will be at mean position present as a straight line.
PHXI15:WAVES
354516
The equation \(y=4+2 \sin (6 t-3 x)\) represents a wave motion, then wave speed and amplitude, respectively are
1 Wave speed 1 unit, amplitude 6 units
2 Wave speed 2 units, amplitude 2 units
3 Wave speed 4 units, amplitude \(1 / 2\) unit
4 Wave speed \(1 / 2\) unit, amplitude 5 units
Explanation:
The shape of the wave is shown in the figure From this graph, we see that amplitude of wave is 2 units. Wave speed, \(v = \frac{{{\rm{Coefficient}}\,\,\,{\rm{of}}\,\,t}}{{{\rm{Coefficeint}}\,\,{\rm{of}}\,\,x}} = \frac{6}{3} = 2{\mkern 1mu} units\)
AIIMS - 2009
PHXI15:WAVES
354517
A wave of frequency \(500\;Hz\) has a velocity \(360\;m{s^{ - 1}}\). The phase difference between two displacements at a certain point at time \({10^{ - 3}}\;s\) apart will be
354518
The phase difference between two waves represented by \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(\quad y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) where \(x\) is expressed in metre and \(t\) is expressed in second, is approximately
1 \(1.07\,rad\)
2 \(2.07\,rad\)
3 \(0.5\,rad\)
4 \(1.5\,rad\)
Explanation:
The given waves equations are \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) \(\begin{array}{r}\Rightarrow y_{2}=10^{-6} \sin \left[100 t+(x / 50)+\dfrac{\pi}{2}\right] m \\{\left[\because \sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right]}\end{array}\) Hence, the phase difference between the waves is \(\Delta \phi = \left( {\frac{\pi }{2} - 0.5} \right)rad = \left( {\frac{{3.14}}{2} - 0.5} \right)rad\) \( = (1.57 - 0.5)\,rad = 1.07\,rad\)
354515
At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is
1 Sinusoidal shape with amplitude \(A\)
2 Straight line
3 Sinusoidal shape with amplitude \(A/3\)
4 Sinusoidal shape with amplitude \(A/2\)
Explanation:
Particles have kinetic energy maximum at mean position. So the string will be at mean position present as a straight line.
PHXI15:WAVES
354516
The equation \(y=4+2 \sin (6 t-3 x)\) represents a wave motion, then wave speed and amplitude, respectively are
1 Wave speed 1 unit, amplitude 6 units
2 Wave speed 2 units, amplitude 2 units
3 Wave speed 4 units, amplitude \(1 / 2\) unit
4 Wave speed \(1 / 2\) unit, amplitude 5 units
Explanation:
The shape of the wave is shown in the figure From this graph, we see that amplitude of wave is 2 units. Wave speed, \(v = \frac{{{\rm{Coefficient}}\,\,\,{\rm{of}}\,\,t}}{{{\rm{Coefficeint}}\,\,{\rm{of}}\,\,x}} = \frac{6}{3} = 2{\mkern 1mu} units\)
AIIMS - 2009
PHXI15:WAVES
354517
A wave of frequency \(500\;Hz\) has a velocity \(360\;m{s^{ - 1}}\). The phase difference between two displacements at a certain point at time \({10^{ - 3}}\;s\) apart will be
354518
The phase difference between two waves represented by \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(\quad y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) where \(x\) is expressed in metre and \(t\) is expressed in second, is approximately
1 \(1.07\,rad\)
2 \(2.07\,rad\)
3 \(0.5\,rad\)
4 \(1.5\,rad\)
Explanation:
The given waves equations are \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) \(\begin{array}{r}\Rightarrow y_{2}=10^{-6} \sin \left[100 t+(x / 50)+\dfrac{\pi}{2}\right] m \\{\left[\because \sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right]}\end{array}\) Hence, the phase difference between the waves is \(\Delta \phi = \left( {\frac{\pi }{2} - 0.5} \right)rad = \left( {\frac{{3.14}}{2} - 0.5} \right)rad\) \( = (1.57 - 0.5)\,rad = 1.07\,rad\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
354515
At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is
1 Sinusoidal shape with amplitude \(A\)
2 Straight line
3 Sinusoidal shape with amplitude \(A/3\)
4 Sinusoidal shape with amplitude \(A/2\)
Explanation:
Particles have kinetic energy maximum at mean position. So the string will be at mean position present as a straight line.
PHXI15:WAVES
354516
The equation \(y=4+2 \sin (6 t-3 x)\) represents a wave motion, then wave speed and amplitude, respectively are
1 Wave speed 1 unit, amplitude 6 units
2 Wave speed 2 units, amplitude 2 units
3 Wave speed 4 units, amplitude \(1 / 2\) unit
4 Wave speed \(1 / 2\) unit, amplitude 5 units
Explanation:
The shape of the wave is shown in the figure From this graph, we see that amplitude of wave is 2 units. Wave speed, \(v = \frac{{{\rm{Coefficient}}\,\,\,{\rm{of}}\,\,t}}{{{\rm{Coefficeint}}\,\,{\rm{of}}\,\,x}} = \frac{6}{3} = 2{\mkern 1mu} units\)
AIIMS - 2009
PHXI15:WAVES
354517
A wave of frequency \(500\;Hz\) has a velocity \(360\;m{s^{ - 1}}\). The phase difference between two displacements at a certain point at time \({10^{ - 3}}\;s\) apart will be
354518
The phase difference between two waves represented by \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(\quad y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) where \(x\) is expressed in metre and \(t\) is expressed in second, is approximately
1 \(1.07\,rad\)
2 \(2.07\,rad\)
3 \(0.5\,rad\)
4 \(1.5\,rad\)
Explanation:
The given waves equations are \(y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m\) and \(y_{2}=10^{-6} \cos [100 t+(x / 50)] m\) \(\begin{array}{r}\Rightarrow y_{2}=10^{-6} \sin \left[100 t+(x / 50)+\dfrac{\pi}{2}\right] m \\{\left[\because \sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right]}\end{array}\) Hence, the phase difference between the waves is \(\Delta \phi = \left( {\frac{\pi }{2} - 0.5} \right)rad = \left( {\frac{{3.14}}{2} - 0.5} \right)rad\) \( = (1.57 - 0.5)\,rad = 1.07\,rad\)
