172265
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $\mathrm{t}$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is :
1 $18^{0}$
2 $36^{0}$
3 $54^{0}$
4 $72^{0}$
Explanation:
D Given, Equation of a wave is given, $y=6 \sin 2 \pi(2 t-0.1 x)$ The equation can be written as $y=6 \sin (4 \pi t-0.2 \pi x)$ From the given equation- $\mathrm{k}=0.2 \pi$ $\frac{2 \pi}{\lambda}=0.2 \pi$ $\lambda=10 \mathrm{~mm}$ $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{10} \times 2$ $\Delta \phi=\frac{2 \pi}{5}=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$
Karnataka CET-2008
WAVES
172267
The equation of a transverse wave travelling along positive $x$ axis with amplitude $0.2 \mathrm{~m}$, velocity $360 \mathrm{~m} / \mathrm{sec}$ and wave-length $60 \mathrm{~m}$ can be written as :
1 $y=0.2 \sin \pi\left[6 t+\frac{x}{60}\right]$
2 $y=0.2 \sin \pi\left[6 t-\frac{x}{60}\right]$
3 $y=0.2 \sin 2 \pi\left[6 t-\frac{x}{60}\right]$
4 $y=0.2 \sin 2 \pi\left[6 t+\frac{x}{60}\right]$
Explanation:
C Equation of plane progressive wave moving in positive direction of $\mathrm{x}$ - axis is given by $y=A \sin (\omega t-k x)$ $\omega=\frac{2 \pi}{T}=2 \pi f$ $\mathrm{y}=\mathrm{A} \sin \left(2 \pi \mathrm{ft}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\mathrm{ft}-\frac{\mathrm{x}}{\lambda}\right)$ $\text { Here, } \quad \mathrm{A}=0.2 \mathrm{~m}, \mathrm{v}=360 \mathrm{~ms}^{-1}, \lambda=60 \mathrm{~m}$ $\text { So, } \quad \mathrm{v}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{360}{60}=6 \mathrm{~Hz} \quad \text { (Where } \mathrm{f}=\text { Frequency) }$ $\mathrm{y}=0.2 \sin 2 \pi\left[6 \mathrm{t}-\frac{\mathrm{x}}{60}\right]$
Karnataka CET-2003
WAVES
172270
A sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then
1 $\mathrm{V}=\mathrm{v}$ if $\lambda=\frac{3 \mathrm{~A}}{2 \pi}$
2 $\mathrm{V}=\mathrm{V}$ if $\mathrm{A}=2 \pi \lambda$
3 $\mathrm{V}=\mathrm{v}$ if $\mathrm{A}=\frac{\lambda}{2 \pi}$
4 $\mathrm{V}$ can not be equal to $\mathrm{V}$.
Explanation:
C $\left(\mathrm{v}_{\text {particle }}\right)_{\text {maximum }}=\mathrm{A} \omega=\mathrm{v}$ $\mathrm{A}=\text { Amplitude, } \omega=\text { angular frequency }$ $\\ \mathrm{v}_{\mathrm{wave}}=\mathrm{f} \lambda=\mathrm{V}$ $\mathrm{f}=\text { Frequency }$ Now, when $\mathrm{V}=\mathrm{V}$ $\text { A } \omega=\mathrm{f} \times \lambda$ $\text { A }(2 \pi \mathrm{f})=\mathrm{f} \times \lambda$ $\mathrm{A}=\frac{\lambda}{2 \pi}$
Karnataka CET-2001
WAVES
172273
Equation of progressive wave is $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$
1 its wavelength is 0.2 units
2 it is travelling in the positive $\mathrm{x}$-direction
3 wave velocity is 1.5 units
4 time period of SHM is $1 \mathrm{~s}$
Explanation:
A Given equation is- $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$ Here, $\mathrm{kx}=10 \pi \mathrm{x}$ $\frac{2 \pi}{\lambda}=10 \pi$ $\lambda=0.2 \mathrm{~m}$ And, $\omega \mathrm{t}=11 \pi \mathrm{t}$ $\frac{2 \pi}{\mathrm{T}}=11 \pi$ $\mathrm{T}=\frac{2}{11} \text { second }$ $\text { And } \quad \mathrm{v}=\frac{\lambda}{\mathrm{T}}=\frac{0.2 \times 11}{2}=\frac{2.2}{2}=1.1 \mathrm{~m} / \mathrm{s} \text {. }$ So, option (a) is correct.
172265
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $\mathrm{t}$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is :
1 $18^{0}$
2 $36^{0}$
3 $54^{0}$
4 $72^{0}$
Explanation:
D Given, Equation of a wave is given, $y=6 \sin 2 \pi(2 t-0.1 x)$ The equation can be written as $y=6 \sin (4 \pi t-0.2 \pi x)$ From the given equation- $\mathrm{k}=0.2 \pi$ $\frac{2 \pi}{\lambda}=0.2 \pi$ $\lambda=10 \mathrm{~mm}$ $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{10} \times 2$ $\Delta \phi=\frac{2 \pi}{5}=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$
Karnataka CET-2008
WAVES
172267
The equation of a transverse wave travelling along positive $x$ axis with amplitude $0.2 \mathrm{~m}$, velocity $360 \mathrm{~m} / \mathrm{sec}$ and wave-length $60 \mathrm{~m}$ can be written as :
1 $y=0.2 \sin \pi\left[6 t+\frac{x}{60}\right]$
2 $y=0.2 \sin \pi\left[6 t-\frac{x}{60}\right]$
3 $y=0.2 \sin 2 \pi\left[6 t-\frac{x}{60}\right]$
4 $y=0.2 \sin 2 \pi\left[6 t+\frac{x}{60}\right]$
Explanation:
C Equation of plane progressive wave moving in positive direction of $\mathrm{x}$ - axis is given by $y=A \sin (\omega t-k x)$ $\omega=\frac{2 \pi}{T}=2 \pi f$ $\mathrm{y}=\mathrm{A} \sin \left(2 \pi \mathrm{ft}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\mathrm{ft}-\frac{\mathrm{x}}{\lambda}\right)$ $\text { Here, } \quad \mathrm{A}=0.2 \mathrm{~m}, \mathrm{v}=360 \mathrm{~ms}^{-1}, \lambda=60 \mathrm{~m}$ $\text { So, } \quad \mathrm{v}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{360}{60}=6 \mathrm{~Hz} \quad \text { (Where } \mathrm{f}=\text { Frequency) }$ $\mathrm{y}=0.2 \sin 2 \pi\left[6 \mathrm{t}-\frac{\mathrm{x}}{60}\right]$
Karnataka CET-2003
WAVES
172270
A sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then
1 $\mathrm{V}=\mathrm{v}$ if $\lambda=\frac{3 \mathrm{~A}}{2 \pi}$
2 $\mathrm{V}=\mathrm{V}$ if $\mathrm{A}=2 \pi \lambda$
3 $\mathrm{V}=\mathrm{v}$ if $\mathrm{A}=\frac{\lambda}{2 \pi}$
4 $\mathrm{V}$ can not be equal to $\mathrm{V}$.
Explanation:
C $\left(\mathrm{v}_{\text {particle }}\right)_{\text {maximum }}=\mathrm{A} \omega=\mathrm{v}$ $\mathrm{A}=\text { Amplitude, } \omega=\text { angular frequency }$ $\\ \mathrm{v}_{\mathrm{wave}}=\mathrm{f} \lambda=\mathrm{V}$ $\mathrm{f}=\text { Frequency }$ Now, when $\mathrm{V}=\mathrm{V}$ $\text { A } \omega=\mathrm{f} \times \lambda$ $\text { A }(2 \pi \mathrm{f})=\mathrm{f} \times \lambda$ $\mathrm{A}=\frac{\lambda}{2 \pi}$
Karnataka CET-2001
WAVES
172273
Equation of progressive wave is $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$
1 its wavelength is 0.2 units
2 it is travelling in the positive $\mathrm{x}$-direction
3 wave velocity is 1.5 units
4 time period of SHM is $1 \mathrm{~s}$
Explanation:
A Given equation is- $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$ Here, $\mathrm{kx}=10 \pi \mathrm{x}$ $\frac{2 \pi}{\lambda}=10 \pi$ $\lambda=0.2 \mathrm{~m}$ And, $\omega \mathrm{t}=11 \pi \mathrm{t}$ $\frac{2 \pi}{\mathrm{T}}=11 \pi$ $\mathrm{T}=\frac{2}{11} \text { second }$ $\text { And } \quad \mathrm{v}=\frac{\lambda}{\mathrm{T}}=\frac{0.2 \times 11}{2}=\frac{2.2}{2}=1.1 \mathrm{~m} / \mathrm{s} \text {. }$ So, option (a) is correct.
172265
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $\mathrm{t}$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is :
1 $18^{0}$
2 $36^{0}$
3 $54^{0}$
4 $72^{0}$
Explanation:
D Given, Equation of a wave is given, $y=6 \sin 2 \pi(2 t-0.1 x)$ The equation can be written as $y=6 \sin (4 \pi t-0.2 \pi x)$ From the given equation- $\mathrm{k}=0.2 \pi$ $\frac{2 \pi}{\lambda}=0.2 \pi$ $\lambda=10 \mathrm{~mm}$ $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{10} \times 2$ $\Delta \phi=\frac{2 \pi}{5}=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$
Karnataka CET-2008
WAVES
172267
The equation of a transverse wave travelling along positive $x$ axis with amplitude $0.2 \mathrm{~m}$, velocity $360 \mathrm{~m} / \mathrm{sec}$ and wave-length $60 \mathrm{~m}$ can be written as :
1 $y=0.2 \sin \pi\left[6 t+\frac{x}{60}\right]$
2 $y=0.2 \sin \pi\left[6 t-\frac{x}{60}\right]$
3 $y=0.2 \sin 2 \pi\left[6 t-\frac{x}{60}\right]$
4 $y=0.2 \sin 2 \pi\left[6 t+\frac{x}{60}\right]$
Explanation:
C Equation of plane progressive wave moving in positive direction of $\mathrm{x}$ - axis is given by $y=A \sin (\omega t-k x)$ $\omega=\frac{2 \pi}{T}=2 \pi f$ $\mathrm{y}=\mathrm{A} \sin \left(2 \pi \mathrm{ft}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\mathrm{ft}-\frac{\mathrm{x}}{\lambda}\right)$ $\text { Here, } \quad \mathrm{A}=0.2 \mathrm{~m}, \mathrm{v}=360 \mathrm{~ms}^{-1}, \lambda=60 \mathrm{~m}$ $\text { So, } \quad \mathrm{v}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{360}{60}=6 \mathrm{~Hz} \quad \text { (Where } \mathrm{f}=\text { Frequency) }$ $\mathrm{y}=0.2 \sin 2 \pi\left[6 \mathrm{t}-\frac{\mathrm{x}}{60}\right]$
Karnataka CET-2003
WAVES
172270
A sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then
1 $\mathrm{V}=\mathrm{v}$ if $\lambda=\frac{3 \mathrm{~A}}{2 \pi}$
2 $\mathrm{V}=\mathrm{V}$ if $\mathrm{A}=2 \pi \lambda$
3 $\mathrm{V}=\mathrm{v}$ if $\mathrm{A}=\frac{\lambda}{2 \pi}$
4 $\mathrm{V}$ can not be equal to $\mathrm{V}$.
Explanation:
C $\left(\mathrm{v}_{\text {particle }}\right)_{\text {maximum }}=\mathrm{A} \omega=\mathrm{v}$ $\mathrm{A}=\text { Amplitude, } \omega=\text { angular frequency }$ $\\ \mathrm{v}_{\mathrm{wave}}=\mathrm{f} \lambda=\mathrm{V}$ $\mathrm{f}=\text { Frequency }$ Now, when $\mathrm{V}=\mathrm{V}$ $\text { A } \omega=\mathrm{f} \times \lambda$ $\text { A }(2 \pi \mathrm{f})=\mathrm{f} \times \lambda$ $\mathrm{A}=\frac{\lambda}{2 \pi}$
Karnataka CET-2001
WAVES
172273
Equation of progressive wave is $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$
1 its wavelength is 0.2 units
2 it is travelling in the positive $\mathrm{x}$-direction
3 wave velocity is 1.5 units
4 time period of SHM is $1 \mathrm{~s}$
Explanation:
A Given equation is- $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$ Here, $\mathrm{kx}=10 \pi \mathrm{x}$ $\frac{2 \pi}{\lambda}=10 \pi$ $\lambda=0.2 \mathrm{~m}$ And, $\omega \mathrm{t}=11 \pi \mathrm{t}$ $\frac{2 \pi}{\mathrm{T}}=11 \pi$ $\mathrm{T}=\frac{2}{11} \text { second }$ $\text { And } \quad \mathrm{v}=\frac{\lambda}{\mathrm{T}}=\frac{0.2 \times 11}{2}=\frac{2.2}{2}=1.1 \mathrm{~m} / \mathrm{s} \text {. }$ So, option (a) is correct.
172265
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $\mathrm{t}$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is :
1 $18^{0}$
2 $36^{0}$
3 $54^{0}$
4 $72^{0}$
Explanation:
D Given, Equation of a wave is given, $y=6 \sin 2 \pi(2 t-0.1 x)$ The equation can be written as $y=6 \sin (4 \pi t-0.2 \pi x)$ From the given equation- $\mathrm{k}=0.2 \pi$ $\frac{2 \pi}{\lambda}=0.2 \pi$ $\lambda=10 \mathrm{~mm}$ $\Delta \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{10} \times 2$ $\Delta \phi=\frac{2 \pi}{5}=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$
Karnataka CET-2008
WAVES
172267
The equation of a transverse wave travelling along positive $x$ axis with amplitude $0.2 \mathrm{~m}$, velocity $360 \mathrm{~m} / \mathrm{sec}$ and wave-length $60 \mathrm{~m}$ can be written as :
1 $y=0.2 \sin \pi\left[6 t+\frac{x}{60}\right]$
2 $y=0.2 \sin \pi\left[6 t-\frac{x}{60}\right]$
3 $y=0.2 \sin 2 \pi\left[6 t-\frac{x}{60}\right]$
4 $y=0.2 \sin 2 \pi\left[6 t+\frac{x}{60}\right]$
Explanation:
C Equation of plane progressive wave moving in positive direction of $\mathrm{x}$ - axis is given by $y=A \sin (\omega t-k x)$ $\omega=\frac{2 \pi}{T}=2 \pi f$ $\mathrm{y}=\mathrm{A} \sin \left(2 \pi \mathrm{ft}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\mathrm{ft}-\frac{\mathrm{x}}{\lambda}\right)$ $\text { Here, } \quad \mathrm{A}=0.2 \mathrm{~m}, \mathrm{v}=360 \mathrm{~ms}^{-1}, \lambda=60 \mathrm{~m}$ $\text { So, } \quad \mathrm{v}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{360}{60}=6 \mathrm{~Hz} \quad \text { (Where } \mathrm{f}=\text { Frequency) }$ $\mathrm{y}=0.2 \sin 2 \pi\left[6 \mathrm{t}-\frac{\mathrm{x}}{60}\right]$
Karnataka CET-2003
WAVES
172270
A sine wave has an amplitude $A$ and wavelength $\lambda$. Let $V$ be the wave velocity and $v$ be the maximum velocity of a particle in the medium. Then
1 $\mathrm{V}=\mathrm{v}$ if $\lambda=\frac{3 \mathrm{~A}}{2 \pi}$
2 $\mathrm{V}=\mathrm{V}$ if $\mathrm{A}=2 \pi \lambda$
3 $\mathrm{V}=\mathrm{v}$ if $\mathrm{A}=\frac{\lambda}{2 \pi}$
4 $\mathrm{V}$ can not be equal to $\mathrm{V}$.
Explanation:
C $\left(\mathrm{v}_{\text {particle }}\right)_{\text {maximum }}=\mathrm{A} \omega=\mathrm{v}$ $\mathrm{A}=\text { Amplitude, } \omega=\text { angular frequency }$ $\\ \mathrm{v}_{\mathrm{wave}}=\mathrm{f} \lambda=\mathrm{V}$ $\mathrm{f}=\text { Frequency }$ Now, when $\mathrm{V}=\mathrm{V}$ $\text { A } \omega=\mathrm{f} \times \lambda$ $\text { A }(2 \pi \mathrm{f})=\mathrm{f} \times \lambda$ $\mathrm{A}=\frac{\lambda}{2 \pi}$
Karnataka CET-2001
WAVES
172273
Equation of progressive wave is $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$
1 its wavelength is 0.2 units
2 it is travelling in the positive $\mathrm{x}$-direction
3 wave velocity is 1.5 units
4 time period of SHM is $1 \mathrm{~s}$
Explanation:
A Given equation is- $y=A \sin \left(10 \pi x+11 \pi t+\frac{\pi}{3}\right)$ Here, $\mathrm{kx}=10 \pi \mathrm{x}$ $\frac{2 \pi}{\lambda}=10 \pi$ $\lambda=0.2 \mathrm{~m}$ And, $\omega \mathrm{t}=11 \pi \mathrm{t}$ $\frac{2 \pi}{\mathrm{T}}=11 \pi$ $\mathrm{T}=\frac{2}{11} \text { second }$ $\text { And } \quad \mathrm{v}=\frac{\lambda}{\mathrm{T}}=\frac{0.2 \times 11}{2}=\frac{2.2}{2}=1.1 \mathrm{~m} / \mathrm{s} \text {. }$ So, option (a) is correct.
