CBSE 11th Standard Physics Subject Waves Ncert Exemplar 5 Mark Questions 2021
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CBSE 11th Standard Physics Subject Waves Ncert Exemplar 5 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Physics
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For the harmonic travelling wave \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \) , where, x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of 0.5 m.
(a) -
For the harmonic travelling wave \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \) , where, x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of \(\frac { \lambda }{2 } \).
(a) -
For the harmonic travelling wave \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \) , where, x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of \(\frac { 3\lambda }{ 4 } \)(at a given instant in time)?
(a) -
For the harmonic travelling wave \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \), where, x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of What is the phase difference between the oscillation of a particle located at x=100 cm, at t=Ts and t = 5s?
(a) -
In the given progressive wave, \(y=5sin(100\pi t-0.4\pi x)\)
Where, y and x are in m,t is in seconds. What is the
(i) amplitude
(ii) wavelength
(iii) frequency
(iv) wave velocity
(v) particle velocity amplitude?(a)
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CBSE 11th Standard Physics Subject Waves Ncert Exemplar 5 Mark Questions 2021 Answer Keys
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Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
\( y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\( k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\phi =2\pi \times 3.5=7\pi rad\)
\({ \phi }_{ 2 }=\frac { 2\pi }{ \lambda } \times x=\frac { 2\pi }{ 1.25 } \times 0.5=0.8\pi rad\). -
Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
\(\\ y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\(k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
When x = \(\frac { \lambda }{2 } \)
\({ \phi }_{ 3 }=\frac { 2\pi }{ \lambda } \times \lambda /2=\pi rad\). -
Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right)\)
\( y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\quad T=0.1s\)
\(\\ k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
When x = \(\frac { 3\lambda }{ 4 } \); \({ \phi }_{ 4 }=\frac { 2\pi }{ \lambda } \times \frac { 3\lambda }{ 4 } =\frac { 3\pi }{ 2 } rad\). -
Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)
\(y=a\cos { \left( \omega t-kx+\phi \right) } \)
\( a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\( k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
\(At\ t=T;\phi =\frac { 2\pi }{ T } =\frac { 2\pi }{ 0.1 } =20\pi rad\)
\(\\ and\ at\ t=5s;{ \phi }^{ ' }=\frac { 2\pi }{ 0.1 } \times 5.100\pi rad\)
\(\therefore phase\ difference\ { \phi }^{ ' }-\ \phi =100\pi rad-20\pi rad=8020\pi rad\). -
Comparing with the standard from the equation
Y = asin\(\left[ \omega t-kx \right] \)
(i) Amplitude, a = 5m
(ii) \(\omega =\frac { 2\pi }{ T } =100\pi \)
\(\\ K=\frac { 2\pi }{ \lambda } =0.4\pi \Rightarrow \lambda =\frac { 2 }{ 0.4 } =\frac { 20 }{ 4 } =5m\)
(iii) \(\omega =2\pi v=100\pi v=50Hz\)
(iv) Wave velocity, v = \(v\lambda \) = 50 x 5 = 250m/s
(v) Particle velocity = \(\frac { dy }{ dx } =a\omega cos(\omega t-kx\)
\(\left( \frac { dy }{ dx } \right) _{ max }=a\omega =5\times 100\pi m/s\)
