172274
$y=25 \cos (2 \pi t-\pi x)$ is the wave equation. Then the amplitude and frequency are respectively
1 100,25
2 200,25
3 25,100
4 $25,1.00$
Explanation:
D Simple Harmonic Motion is represented by, $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx})$ Given that, $\mathrm{y}=25 \cos (2 \pi \mathrm{t}-\pi \mathrm{x})$ Comparing equation (i) and (ii) we get, $A=25 \text { and } \omega=2 \pi$ As we know that frequency is, $\mathrm{f}=\frac{\omega}{2 \pi}=\frac{2 \pi}{2 \pi}=1 \mathrm{~Hz}$
J and K CET- 1999
WAVES
172278
The equation of a wave is $y=5$ $\sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum velocity of the wave particle will be
1 $1 \mathrm{~ms}^{-1}$
2 $2 \mathrm{~ms}^{-1}$
3 $1.5 \mathrm{~ms}^{-1}$
4 $1.25 \mathrm{~ms}^{-1}$
Explanation:
D Equation of wave $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ On comparing the equation with $\mathrm{y}=\mathrm{A} \sin \left(\omega \mathrm{t}-\frac{2 \pi}{\lambda} \mathrm{x}\right)$ $\mathrm{A}=5 \mathrm{~cm}$ $\omega=\frac{1}{0.04}=\frac{100}{4}$ $\omega=25 \mathrm{rad}$ Maximum velocity of the wave, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\text {max }}=5 \times 25$ $\mathrm{v}_{\text {max }}=125 \mathrm{~cm} / \mathrm{s}$ $\mathrm{v}_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}$
UPSEE - 2014
WAVES
172279
If a wave travelling in positive $x$-direction with $A=0.2 \mathrm{~m}$, velocity $=360 \mathrm{~m} / \mathrm{s}$ and $\lambda=60 \mathrm{~m}$, then correct expression for the wave is
C Standard equation of wave travelling in Positive $\mathrm{X}$ - direction, $y=A \sin [\omega t-k x]$ $y=A \sin \left[2 \pi\left(f t-\frac{x}{\lambda}\right)\right]$ Given that, $\mathrm{A}=0.2 \mathrm{~m}$ $\text { Velocity } \mathrm{v}=360 \mathrm{~m} / \mathrm{s} \text { and } \mathrm{f}=\text { frequency }$ $\lambda=60 \mathrm{~m}$ $\text { We know that, }$ $\because \mathrm{v}=\mathrm{f} \lambda$ $360=\mathrm{f} \times 60$ $\mathrm{f}=6$ Putting value of $f, A$ and $\lambda$ in equation (i), we get- $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
Manipal UGET-2016
WAVES
172280
The equation of progressive wave is $y=4 \sin \left(4 \pi t-0.04 x+\frac{\pi}{3}\right)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is
1 $100 \pi \mathrm{m} / \mathrm{s}$
2 $50 \pi \mathrm{m} / \mathrm{s}$
3 $25 \pi \mathrm{m} / \mathrm{s}$
4 $\pi \mathrm{m} / \mathrm{s}$
Explanation:
A $\mathrm{y}=4 \sin \left(4 \pi \mathrm{t}-0.04 \mathrm{x}+\frac{\pi}{3}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)$ Comparing equation (i) with equation (ii) we get- $\mathrm{A}=4, \omega=4 \pi, \mathrm{k}=0.04, \phi=\frac{\pi}{3}$ So velocity of the wave, $\mathrm{v}=\frac{\omega}{\mathrm{k}}$ Putting the value of $\omega$ and $k$ $\mathrm{v}=\frac{4 \pi}{0.04}$ $\mathrm{v}=\frac{4 \pi \times 100}{4}$ $\mathrm{v}=100 \pi \mathrm{m} / \mathrm{s}$
WAVES
172281
A plane progressive wave is given by $y=2$ cos6. $284(330 t-x)$. What is the period of the wave?
1 $\frac{1}{330} \mathrm{~s}$
2 $2 \pi \times 330 \mathrm{~s}$
3 $(2 \pi \times 330)^{-1} \mathrm{~s}$
4 $\frac{6.284}{330} \mathrm{~s}$
Explanation:
A The given equation of a plane progressive wave is $y=2 \cos 6.284(330 t-x)$ $=2 \cos 2 \pi(330 t-x)$ The standard equation of a plane progressive wave is $y=A \cos 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$ On comparing (i) and (ii), we get- $\frac{1}{\mathrm{~T}}=330$ or $\mathrm{T}=\frac{1}{330} \mathrm{~s}$
172274
$y=25 \cos (2 \pi t-\pi x)$ is the wave equation. Then the amplitude and frequency are respectively
1 100,25
2 200,25
3 25,100
4 $25,1.00$
Explanation:
D Simple Harmonic Motion is represented by, $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx})$ Given that, $\mathrm{y}=25 \cos (2 \pi \mathrm{t}-\pi \mathrm{x})$ Comparing equation (i) and (ii) we get, $A=25 \text { and } \omega=2 \pi$ As we know that frequency is, $\mathrm{f}=\frac{\omega}{2 \pi}=\frac{2 \pi}{2 \pi}=1 \mathrm{~Hz}$
J and K CET- 1999
WAVES
172278
The equation of a wave is $y=5$ $\sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum velocity of the wave particle will be
1 $1 \mathrm{~ms}^{-1}$
2 $2 \mathrm{~ms}^{-1}$
3 $1.5 \mathrm{~ms}^{-1}$
4 $1.25 \mathrm{~ms}^{-1}$
Explanation:
D Equation of wave $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ On comparing the equation with $\mathrm{y}=\mathrm{A} \sin \left(\omega \mathrm{t}-\frac{2 \pi}{\lambda} \mathrm{x}\right)$ $\mathrm{A}=5 \mathrm{~cm}$ $\omega=\frac{1}{0.04}=\frac{100}{4}$ $\omega=25 \mathrm{rad}$ Maximum velocity of the wave, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\text {max }}=5 \times 25$ $\mathrm{v}_{\text {max }}=125 \mathrm{~cm} / \mathrm{s}$ $\mathrm{v}_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}$
UPSEE - 2014
WAVES
172279
If a wave travelling in positive $x$-direction with $A=0.2 \mathrm{~m}$, velocity $=360 \mathrm{~m} / \mathrm{s}$ and $\lambda=60 \mathrm{~m}$, then correct expression for the wave is
C Standard equation of wave travelling in Positive $\mathrm{X}$ - direction, $y=A \sin [\omega t-k x]$ $y=A \sin \left[2 \pi\left(f t-\frac{x}{\lambda}\right)\right]$ Given that, $\mathrm{A}=0.2 \mathrm{~m}$ $\text { Velocity } \mathrm{v}=360 \mathrm{~m} / \mathrm{s} \text { and } \mathrm{f}=\text { frequency }$ $\lambda=60 \mathrm{~m}$ $\text { We know that, }$ $\because \mathrm{v}=\mathrm{f} \lambda$ $360=\mathrm{f} \times 60$ $\mathrm{f}=6$ Putting value of $f, A$ and $\lambda$ in equation (i), we get- $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
Manipal UGET-2016
WAVES
172280
The equation of progressive wave is $y=4 \sin \left(4 \pi t-0.04 x+\frac{\pi}{3}\right)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is
1 $100 \pi \mathrm{m} / \mathrm{s}$
2 $50 \pi \mathrm{m} / \mathrm{s}$
3 $25 \pi \mathrm{m} / \mathrm{s}$
4 $\pi \mathrm{m} / \mathrm{s}$
Explanation:
A $\mathrm{y}=4 \sin \left(4 \pi \mathrm{t}-0.04 \mathrm{x}+\frac{\pi}{3}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)$ Comparing equation (i) with equation (ii) we get- $\mathrm{A}=4, \omega=4 \pi, \mathrm{k}=0.04, \phi=\frac{\pi}{3}$ So velocity of the wave, $\mathrm{v}=\frac{\omega}{\mathrm{k}}$ Putting the value of $\omega$ and $k$ $\mathrm{v}=\frac{4 \pi}{0.04}$ $\mathrm{v}=\frac{4 \pi \times 100}{4}$ $\mathrm{v}=100 \pi \mathrm{m} / \mathrm{s}$
WAVES
172281
A plane progressive wave is given by $y=2$ cos6. $284(330 t-x)$. What is the period of the wave?
1 $\frac{1}{330} \mathrm{~s}$
2 $2 \pi \times 330 \mathrm{~s}$
3 $(2 \pi \times 330)^{-1} \mathrm{~s}$
4 $\frac{6.284}{330} \mathrm{~s}$
Explanation:
A The given equation of a plane progressive wave is $y=2 \cos 6.284(330 t-x)$ $=2 \cos 2 \pi(330 t-x)$ The standard equation of a plane progressive wave is $y=A \cos 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$ On comparing (i) and (ii), we get- $\frac{1}{\mathrm{~T}}=330$ or $\mathrm{T}=\frac{1}{330} \mathrm{~s}$
172274
$y=25 \cos (2 \pi t-\pi x)$ is the wave equation. Then the amplitude and frequency are respectively
1 100,25
2 200,25
3 25,100
4 $25,1.00$
Explanation:
D Simple Harmonic Motion is represented by, $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx})$ Given that, $\mathrm{y}=25 \cos (2 \pi \mathrm{t}-\pi \mathrm{x})$ Comparing equation (i) and (ii) we get, $A=25 \text { and } \omega=2 \pi$ As we know that frequency is, $\mathrm{f}=\frac{\omega}{2 \pi}=\frac{2 \pi}{2 \pi}=1 \mathrm{~Hz}$
J and K CET- 1999
WAVES
172278
The equation of a wave is $y=5$ $\sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum velocity of the wave particle will be
1 $1 \mathrm{~ms}^{-1}$
2 $2 \mathrm{~ms}^{-1}$
3 $1.5 \mathrm{~ms}^{-1}$
4 $1.25 \mathrm{~ms}^{-1}$
Explanation:
D Equation of wave $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ On comparing the equation with $\mathrm{y}=\mathrm{A} \sin \left(\omega \mathrm{t}-\frac{2 \pi}{\lambda} \mathrm{x}\right)$ $\mathrm{A}=5 \mathrm{~cm}$ $\omega=\frac{1}{0.04}=\frac{100}{4}$ $\omega=25 \mathrm{rad}$ Maximum velocity of the wave, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\text {max }}=5 \times 25$ $\mathrm{v}_{\text {max }}=125 \mathrm{~cm} / \mathrm{s}$ $\mathrm{v}_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}$
UPSEE - 2014
WAVES
172279
If a wave travelling in positive $x$-direction with $A=0.2 \mathrm{~m}$, velocity $=360 \mathrm{~m} / \mathrm{s}$ and $\lambda=60 \mathrm{~m}$, then correct expression for the wave is
C Standard equation of wave travelling in Positive $\mathrm{X}$ - direction, $y=A \sin [\omega t-k x]$ $y=A \sin \left[2 \pi\left(f t-\frac{x}{\lambda}\right)\right]$ Given that, $\mathrm{A}=0.2 \mathrm{~m}$ $\text { Velocity } \mathrm{v}=360 \mathrm{~m} / \mathrm{s} \text { and } \mathrm{f}=\text { frequency }$ $\lambda=60 \mathrm{~m}$ $\text { We know that, }$ $\because \mathrm{v}=\mathrm{f} \lambda$ $360=\mathrm{f} \times 60$ $\mathrm{f}=6$ Putting value of $f, A$ and $\lambda$ in equation (i), we get- $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
Manipal UGET-2016
WAVES
172280
The equation of progressive wave is $y=4 \sin \left(4 \pi t-0.04 x+\frac{\pi}{3}\right)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is
1 $100 \pi \mathrm{m} / \mathrm{s}$
2 $50 \pi \mathrm{m} / \mathrm{s}$
3 $25 \pi \mathrm{m} / \mathrm{s}$
4 $\pi \mathrm{m} / \mathrm{s}$
Explanation:
A $\mathrm{y}=4 \sin \left(4 \pi \mathrm{t}-0.04 \mathrm{x}+\frac{\pi}{3}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)$ Comparing equation (i) with equation (ii) we get- $\mathrm{A}=4, \omega=4 \pi, \mathrm{k}=0.04, \phi=\frac{\pi}{3}$ So velocity of the wave, $\mathrm{v}=\frac{\omega}{\mathrm{k}}$ Putting the value of $\omega$ and $k$ $\mathrm{v}=\frac{4 \pi}{0.04}$ $\mathrm{v}=\frac{4 \pi \times 100}{4}$ $\mathrm{v}=100 \pi \mathrm{m} / \mathrm{s}$
WAVES
172281
A plane progressive wave is given by $y=2$ cos6. $284(330 t-x)$. What is the period of the wave?
1 $\frac{1}{330} \mathrm{~s}$
2 $2 \pi \times 330 \mathrm{~s}$
3 $(2 \pi \times 330)^{-1} \mathrm{~s}$
4 $\frac{6.284}{330} \mathrm{~s}$
Explanation:
A The given equation of a plane progressive wave is $y=2 \cos 6.284(330 t-x)$ $=2 \cos 2 \pi(330 t-x)$ The standard equation of a plane progressive wave is $y=A \cos 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$ On comparing (i) and (ii), we get- $\frac{1}{\mathrm{~T}}=330$ or $\mathrm{T}=\frac{1}{330} \mathrm{~s}$
172274
$y=25 \cos (2 \pi t-\pi x)$ is the wave equation. Then the amplitude and frequency are respectively
1 100,25
2 200,25
3 25,100
4 $25,1.00$
Explanation:
D Simple Harmonic Motion is represented by, $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx})$ Given that, $\mathrm{y}=25 \cos (2 \pi \mathrm{t}-\pi \mathrm{x})$ Comparing equation (i) and (ii) we get, $A=25 \text { and } \omega=2 \pi$ As we know that frequency is, $\mathrm{f}=\frac{\omega}{2 \pi}=\frac{2 \pi}{2 \pi}=1 \mathrm{~Hz}$
J and K CET- 1999
WAVES
172278
The equation of a wave is $y=5$ $\sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum velocity of the wave particle will be
1 $1 \mathrm{~ms}^{-1}$
2 $2 \mathrm{~ms}^{-1}$
3 $1.5 \mathrm{~ms}^{-1}$
4 $1.25 \mathrm{~ms}^{-1}$
Explanation:
D Equation of wave $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ On comparing the equation with $\mathrm{y}=\mathrm{A} \sin \left(\omega \mathrm{t}-\frac{2 \pi}{\lambda} \mathrm{x}\right)$ $\mathrm{A}=5 \mathrm{~cm}$ $\omega=\frac{1}{0.04}=\frac{100}{4}$ $\omega=25 \mathrm{rad}$ Maximum velocity of the wave, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\text {max }}=5 \times 25$ $\mathrm{v}_{\text {max }}=125 \mathrm{~cm} / \mathrm{s}$ $\mathrm{v}_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}$
UPSEE - 2014
WAVES
172279
If a wave travelling in positive $x$-direction with $A=0.2 \mathrm{~m}$, velocity $=360 \mathrm{~m} / \mathrm{s}$ and $\lambda=60 \mathrm{~m}$, then correct expression for the wave is
C Standard equation of wave travelling in Positive $\mathrm{X}$ - direction, $y=A \sin [\omega t-k x]$ $y=A \sin \left[2 \pi\left(f t-\frac{x}{\lambda}\right)\right]$ Given that, $\mathrm{A}=0.2 \mathrm{~m}$ $\text { Velocity } \mathrm{v}=360 \mathrm{~m} / \mathrm{s} \text { and } \mathrm{f}=\text { frequency }$ $\lambda=60 \mathrm{~m}$ $\text { We know that, }$ $\because \mathrm{v}=\mathrm{f} \lambda$ $360=\mathrm{f} \times 60$ $\mathrm{f}=6$ Putting value of $f, A$ and $\lambda$ in equation (i), we get- $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
Manipal UGET-2016
WAVES
172280
The equation of progressive wave is $y=4 \sin \left(4 \pi t-0.04 x+\frac{\pi}{3}\right)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is
1 $100 \pi \mathrm{m} / \mathrm{s}$
2 $50 \pi \mathrm{m} / \mathrm{s}$
3 $25 \pi \mathrm{m} / \mathrm{s}$
4 $\pi \mathrm{m} / \mathrm{s}$
Explanation:
A $\mathrm{y}=4 \sin \left(4 \pi \mathrm{t}-0.04 \mathrm{x}+\frac{\pi}{3}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)$ Comparing equation (i) with equation (ii) we get- $\mathrm{A}=4, \omega=4 \pi, \mathrm{k}=0.04, \phi=\frac{\pi}{3}$ So velocity of the wave, $\mathrm{v}=\frac{\omega}{\mathrm{k}}$ Putting the value of $\omega$ and $k$ $\mathrm{v}=\frac{4 \pi}{0.04}$ $\mathrm{v}=\frac{4 \pi \times 100}{4}$ $\mathrm{v}=100 \pi \mathrm{m} / \mathrm{s}$
WAVES
172281
A plane progressive wave is given by $y=2$ cos6. $284(330 t-x)$. What is the period of the wave?
1 $\frac{1}{330} \mathrm{~s}$
2 $2 \pi \times 330 \mathrm{~s}$
3 $(2 \pi \times 330)^{-1} \mathrm{~s}$
4 $\frac{6.284}{330} \mathrm{~s}$
Explanation:
A The given equation of a plane progressive wave is $y=2 \cos 6.284(330 t-x)$ $=2 \cos 2 \pi(330 t-x)$ The standard equation of a plane progressive wave is $y=A \cos 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$ On comparing (i) and (ii), we get- $\frac{1}{\mathrm{~T}}=330$ or $\mathrm{T}=\frac{1}{330} \mathrm{~s}$
172274
$y=25 \cos (2 \pi t-\pi x)$ is the wave equation. Then the amplitude and frequency are respectively
1 100,25
2 200,25
3 25,100
4 $25,1.00$
Explanation:
D Simple Harmonic Motion is represented by, $\mathrm{y}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{kx})$ Given that, $\mathrm{y}=25 \cos (2 \pi \mathrm{t}-\pi \mathrm{x})$ Comparing equation (i) and (ii) we get, $A=25 \text { and } \omega=2 \pi$ As we know that frequency is, $\mathrm{f}=\frac{\omega}{2 \pi}=\frac{2 \pi}{2 \pi}=1 \mathrm{~Hz}$
J and K CET- 1999
WAVES
172278
The equation of a wave is $y=5$ $\sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum velocity of the wave particle will be
1 $1 \mathrm{~ms}^{-1}$
2 $2 \mathrm{~ms}^{-1}$
3 $1.5 \mathrm{~ms}^{-1}$
4 $1.25 \mathrm{~ms}^{-1}$
Explanation:
D Equation of wave $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$ On comparing the equation with $\mathrm{y}=\mathrm{A} \sin \left(\omega \mathrm{t}-\frac{2 \pi}{\lambda} \mathrm{x}\right)$ $\mathrm{A}=5 \mathrm{~cm}$ $\omega=\frac{1}{0.04}=\frac{100}{4}$ $\omega=25 \mathrm{rad}$ Maximum velocity of the wave, $\mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\text {max }}=5 \times 25$ $\mathrm{v}_{\text {max }}=125 \mathrm{~cm} / \mathrm{s}$ $\mathrm{v}_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}$
UPSEE - 2014
WAVES
172279
If a wave travelling in positive $x$-direction with $A=0.2 \mathrm{~m}$, velocity $=360 \mathrm{~m} / \mathrm{s}$ and $\lambda=60 \mathrm{~m}$, then correct expression for the wave is
C Standard equation of wave travelling in Positive $\mathrm{X}$ - direction, $y=A \sin [\omega t-k x]$ $y=A \sin \left[2 \pi\left(f t-\frac{x}{\lambda}\right)\right]$ Given that, $\mathrm{A}=0.2 \mathrm{~m}$ $\text { Velocity } \mathrm{v}=360 \mathrm{~m} / \mathrm{s} \text { and } \mathrm{f}=\text { frequency }$ $\lambda=60 \mathrm{~m}$ $\text { We know that, }$ $\because \mathrm{v}=\mathrm{f} \lambda$ $360=\mathrm{f} \times 60$ $\mathrm{f}=6$ Putting value of $f, A$ and $\lambda$ in equation (i), we get- $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
Manipal UGET-2016
WAVES
172280
The equation of progressive wave is $y=4 \sin \left(4 \pi t-0.04 x+\frac{\pi}{3}\right)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is
1 $100 \pi \mathrm{m} / \mathrm{s}$
2 $50 \pi \mathrm{m} / \mathrm{s}$
3 $25 \pi \mathrm{m} / \mathrm{s}$
4 $\pi \mathrm{m} / \mathrm{s}$
Explanation:
A $\mathrm{y}=4 \sin \left(4 \pi \mathrm{t}-0.04 \mathrm{x}+\frac{\pi}{3}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}+\phi)$ Comparing equation (i) with equation (ii) we get- $\mathrm{A}=4, \omega=4 \pi, \mathrm{k}=0.04, \phi=\frac{\pi}{3}$ So velocity of the wave, $\mathrm{v}=\frac{\omega}{\mathrm{k}}$ Putting the value of $\omega$ and $k$ $\mathrm{v}=\frac{4 \pi}{0.04}$ $\mathrm{v}=\frac{4 \pi \times 100}{4}$ $\mathrm{v}=100 \pi \mathrm{m} / \mathrm{s}$
WAVES
172281
A plane progressive wave is given by $y=2$ cos6. $284(330 t-x)$. What is the period of the wave?
1 $\frac{1}{330} \mathrm{~s}$
2 $2 \pi \times 330 \mathrm{~s}$
3 $(2 \pi \times 330)^{-1} \mathrm{~s}$
4 $\frac{6.284}{330} \mathrm{~s}$
Explanation:
A The given equation of a plane progressive wave is $y=2 \cos 6.284(330 t-x)$ $=2 \cos 2 \pi(330 t-x)$ The standard equation of a plane progressive wave is $y=A \cos 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$ On comparing (i) and (ii), we get- $\frac{1}{\mathrm{~T}}=330$ or $\mathrm{T}=\frac{1}{330} \mathrm{~s}$
