172229
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(n t-\frac{x}{\lambda}\right)$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
172230
The equation of a SHM is given by $y=3$ $\sin \frac{\pi}{2}(50 t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum wave velocity is
172232
A simple harmonic progressive wave is represented by the equation $y=8 \sin$ $2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x$-direction is
1 $18^{\circ}$
2 $36^{\circ}$
3 $54^{\circ}$
4 $72^{\circ}$
Explanation:
D Given equation of wave $\mathrm{y}=8 \sin 2 \pi(0.1 \mathrm{x}-2 \mathrm{t})$ On comparing the given equation with the equation of progressive wave $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$ Then, $\quad \lambda=\frac{1}{0.1}=10 \mathrm{~cm}$ So, Phase difference- $\Delta \phi =\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}$ $\Delta \phi =\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}$ $\Delta \phi =\frac{2 \pi}{5} \times \frac{180^{\circ}}{\pi}=72^{\circ}$
CG PET- 2004
WAVES
172233
A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 $\mathrm{cm}$, wavelength $1 \mathrm{~m}$ and wave velocity $5 \mathrm{~m} / \mathrm{s}$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}<0$. Find the wave function $y(x, t)$.
C We start a general form for a rightward moving wave, $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ The given amplitude is $\mathrm{A}=2 \mathrm{~cm}=0.02 \mathrm{~m}$ The wavelength is given as $\lambda=1 \mathrm{~m}$ Wave number $(\mathrm{k})=2 \pi / \lambda=2 \pi \mathrm{m}^{-1}$ Angular frequency, $\omega=\mathrm{vk}=5 \times 2 \pi=10 \pi \mathrm{rad} / \mathrm{s}$ $\text { If } \quad y=0 \text { and } \frac{d y}{d t}<0$ Then, $\quad \phi=2 \mathrm{n} \pi$, where $\mathrm{n}=0,2,4,6 \ldots$. Therefore, by putting value of $\omega, \mathrm{k}, \phi$ we get- $\mathrm{y}(\mathrm{x}, \mathrm{t})=(0.02 \mathrm{~m}) \sin \left[\left(2 \pi \mathrm{m}^{-1}\right) \mathrm{x}-\left(10 \pi \mathrm{s}^{-1}\right) \mathrm{t}\right] \mathrm{m}$
Manipal UGET-2015
WAVES
172234
The equation of the progressive wave is $y=a \sin 2 \pi\left(n t-\frac{x}{5}\right)$. The ratio of maximum particle velocity to wave velocity is
172229
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(n t-\frac{x}{\lambda}\right)$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
172230
The equation of a SHM is given by $y=3$ $\sin \frac{\pi}{2}(50 t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum wave velocity is
172232
A simple harmonic progressive wave is represented by the equation $y=8 \sin$ $2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x$-direction is
1 $18^{\circ}$
2 $36^{\circ}$
3 $54^{\circ}$
4 $72^{\circ}$
Explanation:
D Given equation of wave $\mathrm{y}=8 \sin 2 \pi(0.1 \mathrm{x}-2 \mathrm{t})$ On comparing the given equation with the equation of progressive wave $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$ Then, $\quad \lambda=\frac{1}{0.1}=10 \mathrm{~cm}$ So, Phase difference- $\Delta \phi =\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}$ $\Delta \phi =\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}$ $\Delta \phi =\frac{2 \pi}{5} \times \frac{180^{\circ}}{\pi}=72^{\circ}$
CG PET- 2004
WAVES
172233
A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 $\mathrm{cm}$, wavelength $1 \mathrm{~m}$ and wave velocity $5 \mathrm{~m} / \mathrm{s}$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}<0$. Find the wave function $y(x, t)$.
C We start a general form for a rightward moving wave, $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ The given amplitude is $\mathrm{A}=2 \mathrm{~cm}=0.02 \mathrm{~m}$ The wavelength is given as $\lambda=1 \mathrm{~m}$ Wave number $(\mathrm{k})=2 \pi / \lambda=2 \pi \mathrm{m}^{-1}$ Angular frequency, $\omega=\mathrm{vk}=5 \times 2 \pi=10 \pi \mathrm{rad} / \mathrm{s}$ $\text { If } \quad y=0 \text { and } \frac{d y}{d t}<0$ Then, $\quad \phi=2 \mathrm{n} \pi$, where $\mathrm{n}=0,2,4,6 \ldots$. Therefore, by putting value of $\omega, \mathrm{k}, \phi$ we get- $\mathrm{y}(\mathrm{x}, \mathrm{t})=(0.02 \mathrm{~m}) \sin \left[\left(2 \pi \mathrm{m}^{-1}\right) \mathrm{x}-\left(10 \pi \mathrm{s}^{-1}\right) \mathrm{t}\right] \mathrm{m}$
Manipal UGET-2015
WAVES
172234
The equation of the progressive wave is $y=a \sin 2 \pi\left(n t-\frac{x}{5}\right)$. The ratio of maximum particle velocity to wave velocity is
172229
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(n t-\frac{x}{\lambda}\right)$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
172230
The equation of a SHM is given by $y=3$ $\sin \frac{\pi}{2}(50 t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum wave velocity is
172232
A simple harmonic progressive wave is represented by the equation $y=8 \sin$ $2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x$-direction is
1 $18^{\circ}$
2 $36^{\circ}$
3 $54^{\circ}$
4 $72^{\circ}$
Explanation:
D Given equation of wave $\mathrm{y}=8 \sin 2 \pi(0.1 \mathrm{x}-2 \mathrm{t})$ On comparing the given equation with the equation of progressive wave $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$ Then, $\quad \lambda=\frac{1}{0.1}=10 \mathrm{~cm}$ So, Phase difference- $\Delta \phi =\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}$ $\Delta \phi =\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}$ $\Delta \phi =\frac{2 \pi}{5} \times \frac{180^{\circ}}{\pi}=72^{\circ}$
CG PET- 2004
WAVES
172233
A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 $\mathrm{cm}$, wavelength $1 \mathrm{~m}$ and wave velocity $5 \mathrm{~m} / \mathrm{s}$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}<0$. Find the wave function $y(x, t)$.
C We start a general form for a rightward moving wave, $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ The given amplitude is $\mathrm{A}=2 \mathrm{~cm}=0.02 \mathrm{~m}$ The wavelength is given as $\lambda=1 \mathrm{~m}$ Wave number $(\mathrm{k})=2 \pi / \lambda=2 \pi \mathrm{m}^{-1}$ Angular frequency, $\omega=\mathrm{vk}=5 \times 2 \pi=10 \pi \mathrm{rad} / \mathrm{s}$ $\text { If } \quad y=0 \text { and } \frac{d y}{d t}<0$ Then, $\quad \phi=2 \mathrm{n} \pi$, where $\mathrm{n}=0,2,4,6 \ldots$. Therefore, by putting value of $\omega, \mathrm{k}, \phi$ we get- $\mathrm{y}(\mathrm{x}, \mathrm{t})=(0.02 \mathrm{~m}) \sin \left[\left(2 \pi \mathrm{m}^{-1}\right) \mathrm{x}-\left(10 \pi \mathrm{s}^{-1}\right) \mathrm{t}\right] \mathrm{m}$
Manipal UGET-2015
WAVES
172234
The equation of the progressive wave is $y=a \sin 2 \pi\left(n t-\frac{x}{5}\right)$. The ratio of maximum particle velocity to wave velocity is
172229
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(n t-\frac{x}{\lambda}\right)$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
172230
The equation of a SHM is given by $y=3$ $\sin \frac{\pi}{2}(50 t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum wave velocity is
172232
A simple harmonic progressive wave is represented by the equation $y=8 \sin$ $2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x$-direction is
1 $18^{\circ}$
2 $36^{\circ}$
3 $54^{\circ}$
4 $72^{\circ}$
Explanation:
D Given equation of wave $\mathrm{y}=8 \sin 2 \pi(0.1 \mathrm{x}-2 \mathrm{t})$ On comparing the given equation with the equation of progressive wave $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$ Then, $\quad \lambda=\frac{1}{0.1}=10 \mathrm{~cm}$ So, Phase difference- $\Delta \phi =\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}$ $\Delta \phi =\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}$ $\Delta \phi =\frac{2 \pi}{5} \times \frac{180^{\circ}}{\pi}=72^{\circ}$
CG PET- 2004
WAVES
172233
A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 $\mathrm{cm}$, wavelength $1 \mathrm{~m}$ and wave velocity $5 \mathrm{~m} / \mathrm{s}$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}<0$. Find the wave function $y(x, t)$.
C We start a general form for a rightward moving wave, $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ The given amplitude is $\mathrm{A}=2 \mathrm{~cm}=0.02 \mathrm{~m}$ The wavelength is given as $\lambda=1 \mathrm{~m}$ Wave number $(\mathrm{k})=2 \pi / \lambda=2 \pi \mathrm{m}^{-1}$ Angular frequency, $\omega=\mathrm{vk}=5 \times 2 \pi=10 \pi \mathrm{rad} / \mathrm{s}$ $\text { If } \quad y=0 \text { and } \frac{d y}{d t}<0$ Then, $\quad \phi=2 \mathrm{n} \pi$, where $\mathrm{n}=0,2,4,6 \ldots$. Therefore, by putting value of $\omega, \mathrm{k}, \phi$ we get- $\mathrm{y}(\mathrm{x}, \mathrm{t})=(0.02 \mathrm{~m}) \sin \left[\left(2 \pi \mathrm{m}^{-1}\right) \mathrm{x}-\left(10 \pi \mathrm{s}^{-1}\right) \mathrm{t}\right] \mathrm{m}$
Manipal UGET-2015
WAVES
172234
The equation of the progressive wave is $y=a \sin 2 \pi\left(n t-\frac{x}{5}\right)$. The ratio of maximum particle velocity to wave velocity is
172229
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(n t-\frac{x}{\lambda}\right)$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
172230
The equation of a SHM is given by $y=3$ $\sin \frac{\pi}{2}(50 t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum wave velocity is
172232
A simple harmonic progressive wave is represented by the equation $y=8 \sin$ $2 \pi(0.1 x-2 t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ is in second. At any instant the phase difference between two particles separated by $2.0 \mathrm{~cm}$ in the $x$-direction is
1 $18^{\circ}$
2 $36^{\circ}$
3 $54^{\circ}$
4 $72^{\circ}$
Explanation:
D Given equation of wave $\mathrm{y}=8 \sin 2 \pi(0.1 \mathrm{x}-2 \mathrm{t})$ On comparing the given equation with the equation of progressive wave $\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)$ Then, $\quad \lambda=\frac{1}{0.1}=10 \mathrm{~cm}$ So, Phase difference- $\Delta \phi =\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}$ $\Delta \phi =\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}$ $\Delta \phi =\frac{2 \pi}{5} \times \frac{180^{\circ}}{\pi}=72^{\circ}$
CG PET- 2004
WAVES
172233
A sinusoidal wave travelling in the positive direction on stretched string has amplitude 20 $\mathrm{cm}$, wavelength $1 \mathrm{~m}$ and wave velocity $5 \mathrm{~m} / \mathrm{s}$. At $x=0$ and $t=0$, it is given that $y=0$ and $\frac{d y}{d t}<0$. Find the wave function $y(x, t)$.
C We start a general form for a rightward moving wave, $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ The given amplitude is $\mathrm{A}=2 \mathrm{~cm}=0.02 \mathrm{~m}$ The wavelength is given as $\lambda=1 \mathrm{~m}$ Wave number $(\mathrm{k})=2 \pi / \lambda=2 \pi \mathrm{m}^{-1}$ Angular frequency, $\omega=\mathrm{vk}=5 \times 2 \pi=10 \pi \mathrm{rad} / \mathrm{s}$ $\text { If } \quad y=0 \text { and } \frac{d y}{d t}<0$ Then, $\quad \phi=2 \mathrm{n} \pi$, where $\mathrm{n}=0,2,4,6 \ldots$. Therefore, by putting value of $\omega, \mathrm{k}, \phi$ we get- $\mathrm{y}(\mathrm{x}, \mathrm{t})=(0.02 \mathrm{~m}) \sin \left[\left(2 \pi \mathrm{m}^{-1}\right) \mathrm{x}-\left(10 \pi \mathrm{s}^{-1}\right) \mathrm{t}\right] \mathrm{m}$
Manipal UGET-2015
WAVES
172234
The equation of the progressive wave is $y=a \sin 2 \pi\left(n t-\frac{x}{5}\right)$. The ratio of maximum particle velocity to wave velocity is
