172332
Two progressive waves are represented by the following equations $y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ Find the ratio of their intensities.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{16}$
Explanation:
B The given two Progressive equation $Y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $Y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ From these equation $\mathrm{A}_{1}=10 \mathrm{~m}$ $\mathrm{~A}_{2}=20 \mathrm{~m}$ We know that $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{10}{20}\right)^{2}$ $\mathrm{I}_{1}: \mathrm{I}_{2}=1: 4$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{4}$
AMU-2014
WAVES
172334
A string oscillates according to the equation, $y$ $=\left(\begin{array}{ll}0.50 \mathrm{~cm}\end{array}\right) \sin \left[\frac{\pi x}{3}\right] \cos [40 \pi t]$. Find the distance between nodes.
1 $1.5 \mathrm{~cm}$
2 $2.5 \mathrm{~cm}$
3 $3.0 \mathrm{~cm}$
4 $3.5 \mathrm{~cm}$
Explanation:
C Given equation $\mathrm{y}=0.50 \sin \left[\frac{\pi \mathrm{x}}{3}\right] \cos [40 \pi \mathrm{t}]$ Standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi t}{\lambda}$ On comparing with equation (i) and (ii) $\frac{2 \pi \mathrm{x}}{\lambda}=\frac{\pi \mathrm{x}}{3}$ $\lambda=6 \mathrm{~cm}$ Distance between two nodes $=\frac{\lambda}{2}$ $=\frac{6}{2}$ $=3.0 \mathrm{~cm}$
AMU-2013
WAVES
172335
A transverse wave is represented by the equation $y=2 \sin (30 t-40 x)$ and the measurements of distances are in meters, then the velocity of propagation is
1 $15 \mathrm{~ms}^{-1}$
2 $0.75 \mathrm{~ms}^{-1}$
3 $3.75 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B The wave equation is given by $\mathrm{y}=2 \sin (30 \mathrm{t}-40 \mathrm{x})$ The standard transverse wave is given by $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ On comparing with standard equation $\omega=30$ $k=40$ Velocity $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{40}=0.75 \mathrm{~m} / \mathrm{sec}$
AP EAMCET -2015
WAVES
172336
The displacement of a particle varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is
1 -4
2 4
3 $4 \sqrt{2}$
4 8
Explanation:
C The displacement equation $\mathrm{x}=(4 \cos \pi \mathrm{t}+4 \sin \pi \mathrm{t})$ Let, $\mathrm{R} \cos \phi=4$ $\mathrm{R} \sin \phi=4$ $\mathrm{x}=\mathrm{R} \sin \phi \cos \mathrm{t} \pi \mathrm{t}+\mathrm{R} \cos \phi \sin \pi \mathrm{t}$ $\mathrm{x}=\mathrm{R} \sin (\pi \mathrm{t}+\phi)$ The $\mathrm{R}$ is aptitude $\mathrm{R}^{2} \sin ^{2} \phi+\mathrm{R}^{2} \cos ^{2} \phi=\sqrt{(4)^{2}+(4)^{2}}$ $\mathrm{R}^{2}=\sqrt{16+16}$ $\mathrm{R}=4 \sqrt{2}$ Hence, the amplitudes $4 \sqrt{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172332
Two progressive waves are represented by the following equations $y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ Find the ratio of their intensities.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{16}$
Explanation:
B The given two Progressive equation $Y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $Y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ From these equation $\mathrm{A}_{1}=10 \mathrm{~m}$ $\mathrm{~A}_{2}=20 \mathrm{~m}$ We know that $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{10}{20}\right)^{2}$ $\mathrm{I}_{1}: \mathrm{I}_{2}=1: 4$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{4}$
AMU-2014
WAVES
172334
A string oscillates according to the equation, $y$ $=\left(\begin{array}{ll}0.50 \mathrm{~cm}\end{array}\right) \sin \left[\frac{\pi x}{3}\right] \cos [40 \pi t]$. Find the distance between nodes.
1 $1.5 \mathrm{~cm}$
2 $2.5 \mathrm{~cm}$
3 $3.0 \mathrm{~cm}$
4 $3.5 \mathrm{~cm}$
Explanation:
C Given equation $\mathrm{y}=0.50 \sin \left[\frac{\pi \mathrm{x}}{3}\right] \cos [40 \pi \mathrm{t}]$ Standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi t}{\lambda}$ On comparing with equation (i) and (ii) $\frac{2 \pi \mathrm{x}}{\lambda}=\frac{\pi \mathrm{x}}{3}$ $\lambda=6 \mathrm{~cm}$ Distance between two nodes $=\frac{\lambda}{2}$ $=\frac{6}{2}$ $=3.0 \mathrm{~cm}$
AMU-2013
WAVES
172335
A transverse wave is represented by the equation $y=2 \sin (30 t-40 x)$ and the measurements of distances are in meters, then the velocity of propagation is
1 $15 \mathrm{~ms}^{-1}$
2 $0.75 \mathrm{~ms}^{-1}$
3 $3.75 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B The wave equation is given by $\mathrm{y}=2 \sin (30 \mathrm{t}-40 \mathrm{x})$ The standard transverse wave is given by $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ On comparing with standard equation $\omega=30$ $k=40$ Velocity $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{40}=0.75 \mathrm{~m} / \mathrm{sec}$
AP EAMCET -2015
WAVES
172336
The displacement of a particle varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is
1 -4
2 4
3 $4 \sqrt{2}$
4 8
Explanation:
C The displacement equation $\mathrm{x}=(4 \cos \pi \mathrm{t}+4 \sin \pi \mathrm{t})$ Let, $\mathrm{R} \cos \phi=4$ $\mathrm{R} \sin \phi=4$ $\mathrm{x}=\mathrm{R} \sin \phi \cos \mathrm{t} \pi \mathrm{t}+\mathrm{R} \cos \phi \sin \pi \mathrm{t}$ $\mathrm{x}=\mathrm{R} \sin (\pi \mathrm{t}+\phi)$ The $\mathrm{R}$ is aptitude $\mathrm{R}^{2} \sin ^{2} \phi+\mathrm{R}^{2} \cos ^{2} \phi=\sqrt{(4)^{2}+(4)^{2}}$ $\mathrm{R}^{2}=\sqrt{16+16}$ $\mathrm{R}=4 \sqrt{2}$ Hence, the amplitudes $4 \sqrt{2}$
172332
Two progressive waves are represented by the following equations $y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ Find the ratio of their intensities.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{16}$
Explanation:
B The given two Progressive equation $Y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $Y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ From these equation $\mathrm{A}_{1}=10 \mathrm{~m}$ $\mathrm{~A}_{2}=20 \mathrm{~m}$ We know that $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{10}{20}\right)^{2}$ $\mathrm{I}_{1}: \mathrm{I}_{2}=1: 4$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{4}$
AMU-2014
WAVES
172334
A string oscillates according to the equation, $y$ $=\left(\begin{array}{ll}0.50 \mathrm{~cm}\end{array}\right) \sin \left[\frac{\pi x}{3}\right] \cos [40 \pi t]$. Find the distance between nodes.
1 $1.5 \mathrm{~cm}$
2 $2.5 \mathrm{~cm}$
3 $3.0 \mathrm{~cm}$
4 $3.5 \mathrm{~cm}$
Explanation:
C Given equation $\mathrm{y}=0.50 \sin \left[\frac{\pi \mathrm{x}}{3}\right] \cos [40 \pi \mathrm{t}]$ Standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi t}{\lambda}$ On comparing with equation (i) and (ii) $\frac{2 \pi \mathrm{x}}{\lambda}=\frac{\pi \mathrm{x}}{3}$ $\lambda=6 \mathrm{~cm}$ Distance between two nodes $=\frac{\lambda}{2}$ $=\frac{6}{2}$ $=3.0 \mathrm{~cm}$
AMU-2013
WAVES
172335
A transverse wave is represented by the equation $y=2 \sin (30 t-40 x)$ and the measurements of distances are in meters, then the velocity of propagation is
1 $15 \mathrm{~ms}^{-1}$
2 $0.75 \mathrm{~ms}^{-1}$
3 $3.75 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B The wave equation is given by $\mathrm{y}=2 \sin (30 \mathrm{t}-40 \mathrm{x})$ The standard transverse wave is given by $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ On comparing with standard equation $\omega=30$ $k=40$ Velocity $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{40}=0.75 \mathrm{~m} / \mathrm{sec}$
AP EAMCET -2015
WAVES
172336
The displacement of a particle varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is
1 -4
2 4
3 $4 \sqrt{2}$
4 8
Explanation:
C The displacement equation $\mathrm{x}=(4 \cos \pi \mathrm{t}+4 \sin \pi \mathrm{t})$ Let, $\mathrm{R} \cos \phi=4$ $\mathrm{R} \sin \phi=4$ $\mathrm{x}=\mathrm{R} \sin \phi \cos \mathrm{t} \pi \mathrm{t}+\mathrm{R} \cos \phi \sin \pi \mathrm{t}$ $\mathrm{x}=\mathrm{R} \sin (\pi \mathrm{t}+\phi)$ The $\mathrm{R}$ is aptitude $\mathrm{R}^{2} \sin ^{2} \phi+\mathrm{R}^{2} \cos ^{2} \phi=\sqrt{(4)^{2}+(4)^{2}}$ $\mathrm{R}^{2}=\sqrt{16+16}$ $\mathrm{R}=4 \sqrt{2}$ Hence, the amplitudes $4 \sqrt{2}$
172332
Two progressive waves are represented by the following equations $y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ Find the ratio of their intensities.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{16}$
Explanation:
B The given two Progressive equation $Y_{1}=10 \sin 2 \pi(10 t-0.1 x)$ $Y_{2}=20 \sin 2 \pi(20 t-0.2 x)$ From these equation $\mathrm{A}_{1}=10 \mathrm{~m}$ $\mathrm{~A}_{2}=20 \mathrm{~m}$ We know that $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{10}{20}\right)^{2}$ $\mathrm{I}_{1}: \mathrm{I}_{2}=1: 4$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{4}$
AMU-2014
WAVES
172334
A string oscillates according to the equation, $y$ $=\left(\begin{array}{ll}0.50 \mathrm{~cm}\end{array}\right) \sin \left[\frac{\pi x}{3}\right] \cos [40 \pi t]$. Find the distance between nodes.
1 $1.5 \mathrm{~cm}$
2 $2.5 \mathrm{~cm}$
3 $3.0 \mathrm{~cm}$
4 $3.5 \mathrm{~cm}$
Explanation:
C Given equation $\mathrm{y}=0.50 \sin \left[\frac{\pi \mathrm{x}}{3}\right] \cos [40 \pi \mathrm{t}]$ Standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi t}{\lambda}$ On comparing with equation (i) and (ii) $\frac{2 \pi \mathrm{x}}{\lambda}=\frac{\pi \mathrm{x}}{3}$ $\lambda=6 \mathrm{~cm}$ Distance between two nodes $=\frac{\lambda}{2}$ $=\frac{6}{2}$ $=3.0 \mathrm{~cm}$
AMU-2013
WAVES
172335
A transverse wave is represented by the equation $y=2 \sin (30 t-40 x)$ and the measurements of distances are in meters, then the velocity of propagation is
1 $15 \mathrm{~ms}^{-1}$
2 $0.75 \mathrm{~ms}^{-1}$
3 $3.75 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B The wave equation is given by $\mathrm{y}=2 \sin (30 \mathrm{t}-40 \mathrm{x})$ The standard transverse wave is given by $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ On comparing with standard equation $\omega=30$ $k=40$ Velocity $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{30}{40}=0.75 \mathrm{~m} / \mathrm{sec}$
AP EAMCET -2015
WAVES
172336
The displacement of a particle varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is
1 -4
2 4
3 $4 \sqrt{2}$
4 8
Explanation:
C The displacement equation $\mathrm{x}=(4 \cos \pi \mathrm{t}+4 \sin \pi \mathrm{t})$ Let, $\mathrm{R} \cos \phi=4$ $\mathrm{R} \sin \phi=4$ $\mathrm{x}=\mathrm{R} \sin \phi \cos \mathrm{t} \pi \mathrm{t}+\mathrm{R} \cos \phi \sin \pi \mathrm{t}$ $\mathrm{x}=\mathrm{R} \sin (\pi \mathrm{t}+\phi)$ The $\mathrm{R}$ is aptitude $\mathrm{R}^{2} \sin ^{2} \phi+\mathrm{R}^{2} \cos ^{2} \phi=\sqrt{(4)^{2}+(4)^{2}}$ $\mathrm{R}^{2}=\sqrt{16+16}$ $\mathrm{R}=4 \sqrt{2}$ Hence, the amplitudes $4 \sqrt{2}$