172292
The wavelength of a wave in a medium is $0.5 \mathrm{~m}$. The phase difference between the oscillations at two points in the medium due to this wave is $\frac{\pi}{5}$. What is the minimum distance between these points?
1 $0.05 \mathrm{~m}$
2 $0.1 \mathrm{~m}$
3 $0.25 \mathrm{~m}$
4 $0.15 \mathrm{~m}$
Explanation:
A Given that Wavelength $\lambda=0.5$ Phase change $\Delta \phi=\pi / 5$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ From equation (i) $\pi / 5=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $\Delta \mathrm{x}=\frac{\lambda}{10}$ $=\frac{0.5}{10}=0.05 \mathrm{~m}$
MP PMT-2013
WAVES
172295
If $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right) y \rightarrow m m, t \rightarrow$ second, $\mathbf{x} \rightarrow \mathbf{m}$. For given progressive wave equation, phase difference between two vibrating particle having path difference $3.5 \mathrm{~m}$ would be
1 $\pi / 4$
2 $\pi$
3 $\pi / 3$
4 $\pi / 2$
Explanation:
D Given that $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Comparing the equation (i) and (ii) $\mathrm{A}=5$ $\omega=30 \pi$ $\mathrm{k}=\pi / 7$ Wave length, $\lambda=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \pi \times 7}{\pi}$ $\lambda=14 \mathrm{~m}$ Relation between phase difference and path difference $\Delta \phi =\frac{2 \pi}{\lambda} \times \Delta x$ $=\frac{2 \pi}{14} \times 3.5$ $\Delta \phi =\frac{\pi}{2}$
MP PET-2008
WAVES
172297
$Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B The given equation of a progressive wave is $Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $\mathrm{Y}=3 \sin 2 \pi\left(\frac{\mathrm{t}}{4}-\frac{\mathrm{x}}{8}\right)$ The standard equation of a progressive wave is $\mathrm{Y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ Comparing eq. (i) and (ii) $\mathrm{T}=4 \mathrm{~s} \quad \lambda=8 \mathrm{~m}$ Velocity of wave, $v=\frac{\lambda}{T}$ $\mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{sec}$ Distance travelled by wave in time $\mathrm{t}$ is $\mathrm{S}=\mathrm{vt}$ $\mathrm{S}=2 \times 5 \quad$ (Given, $\mathrm{t}=5 \mathrm{sec}$ ) $\mathrm{S}=10 \mathrm{~m}$
JIPMER-2012
WAVES
172298
The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40$ $\pi \mathrm{t}$, where, $\mathrm{x}$ and $\mathrm{y}$ are in $\mathrm{cm}$ and $\mathrm{t}$ in second. The separation between two adjacent nodes is:
1 $1.5 \mathrm{~cm}$
2 $3 \mathrm{~cm}$
3 $6 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Distance between adjacent nodes is half of wavelength. The standard equation of stationary wave is $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{x}}{\lambda} \times \cos \frac{2 \pi \mathrm{t}}{\lambda}$ Where a is amplitude, $\lambda$ is wavelength. Given equation is $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$ Comparing eq $^{\mathrm{n}}$ (i) and eq ${ }^{\mathrm{n}}$ (ii), we get, $\frac{2 \pi}{\lambda}=\frac{\pi}{3}$ $\therefore \quad \lambda=6 \mathrm{~cm}$ Distance between two adjacent nodes $\Rightarrow \frac{\lambda}{2}=\frac{6}{2}$ $\lambda=3 \mathrm{~cm}$
172292
The wavelength of a wave in a medium is $0.5 \mathrm{~m}$. The phase difference between the oscillations at two points in the medium due to this wave is $\frac{\pi}{5}$. What is the minimum distance between these points?
1 $0.05 \mathrm{~m}$
2 $0.1 \mathrm{~m}$
3 $0.25 \mathrm{~m}$
4 $0.15 \mathrm{~m}$
Explanation:
A Given that Wavelength $\lambda=0.5$ Phase change $\Delta \phi=\pi / 5$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ From equation (i) $\pi / 5=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $\Delta \mathrm{x}=\frac{\lambda}{10}$ $=\frac{0.5}{10}=0.05 \mathrm{~m}$
MP PMT-2013
WAVES
172295
If $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right) y \rightarrow m m, t \rightarrow$ second, $\mathbf{x} \rightarrow \mathbf{m}$. For given progressive wave equation, phase difference between two vibrating particle having path difference $3.5 \mathrm{~m}$ would be
1 $\pi / 4$
2 $\pi$
3 $\pi / 3$
4 $\pi / 2$
Explanation:
D Given that $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Comparing the equation (i) and (ii) $\mathrm{A}=5$ $\omega=30 \pi$ $\mathrm{k}=\pi / 7$ Wave length, $\lambda=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \pi \times 7}{\pi}$ $\lambda=14 \mathrm{~m}$ Relation between phase difference and path difference $\Delta \phi =\frac{2 \pi}{\lambda} \times \Delta x$ $=\frac{2 \pi}{14} \times 3.5$ $\Delta \phi =\frac{\pi}{2}$
MP PET-2008
WAVES
172297
$Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B The given equation of a progressive wave is $Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $\mathrm{Y}=3 \sin 2 \pi\left(\frac{\mathrm{t}}{4}-\frac{\mathrm{x}}{8}\right)$ The standard equation of a progressive wave is $\mathrm{Y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ Comparing eq. (i) and (ii) $\mathrm{T}=4 \mathrm{~s} \quad \lambda=8 \mathrm{~m}$ Velocity of wave, $v=\frac{\lambda}{T}$ $\mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{sec}$ Distance travelled by wave in time $\mathrm{t}$ is $\mathrm{S}=\mathrm{vt}$ $\mathrm{S}=2 \times 5 \quad$ (Given, $\mathrm{t}=5 \mathrm{sec}$ ) $\mathrm{S}=10 \mathrm{~m}$
JIPMER-2012
WAVES
172298
The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40$ $\pi \mathrm{t}$, where, $\mathrm{x}$ and $\mathrm{y}$ are in $\mathrm{cm}$ and $\mathrm{t}$ in second. The separation between two adjacent nodes is:
1 $1.5 \mathrm{~cm}$
2 $3 \mathrm{~cm}$
3 $6 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Distance between adjacent nodes is half of wavelength. The standard equation of stationary wave is $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{x}}{\lambda} \times \cos \frac{2 \pi \mathrm{t}}{\lambda}$ Where a is amplitude, $\lambda$ is wavelength. Given equation is $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$ Comparing eq $^{\mathrm{n}}$ (i) and eq ${ }^{\mathrm{n}}$ (ii), we get, $\frac{2 \pi}{\lambda}=\frac{\pi}{3}$ $\therefore \quad \lambda=6 \mathrm{~cm}$ Distance between two adjacent nodes $\Rightarrow \frac{\lambda}{2}=\frac{6}{2}$ $\lambda=3 \mathrm{~cm}$
172292
The wavelength of a wave in a medium is $0.5 \mathrm{~m}$. The phase difference between the oscillations at two points in the medium due to this wave is $\frac{\pi}{5}$. What is the minimum distance between these points?
1 $0.05 \mathrm{~m}$
2 $0.1 \mathrm{~m}$
3 $0.25 \mathrm{~m}$
4 $0.15 \mathrm{~m}$
Explanation:
A Given that Wavelength $\lambda=0.5$ Phase change $\Delta \phi=\pi / 5$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ From equation (i) $\pi / 5=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $\Delta \mathrm{x}=\frac{\lambda}{10}$ $=\frac{0.5}{10}=0.05 \mathrm{~m}$
MP PMT-2013
WAVES
172295
If $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right) y \rightarrow m m, t \rightarrow$ second, $\mathbf{x} \rightarrow \mathbf{m}$. For given progressive wave equation, phase difference between two vibrating particle having path difference $3.5 \mathrm{~m}$ would be
1 $\pi / 4$
2 $\pi$
3 $\pi / 3$
4 $\pi / 2$
Explanation:
D Given that $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Comparing the equation (i) and (ii) $\mathrm{A}=5$ $\omega=30 \pi$ $\mathrm{k}=\pi / 7$ Wave length, $\lambda=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \pi \times 7}{\pi}$ $\lambda=14 \mathrm{~m}$ Relation between phase difference and path difference $\Delta \phi =\frac{2 \pi}{\lambda} \times \Delta x$ $=\frac{2 \pi}{14} \times 3.5$ $\Delta \phi =\frac{\pi}{2}$
MP PET-2008
WAVES
172297
$Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B The given equation of a progressive wave is $Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $\mathrm{Y}=3 \sin 2 \pi\left(\frac{\mathrm{t}}{4}-\frac{\mathrm{x}}{8}\right)$ The standard equation of a progressive wave is $\mathrm{Y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ Comparing eq. (i) and (ii) $\mathrm{T}=4 \mathrm{~s} \quad \lambda=8 \mathrm{~m}$ Velocity of wave, $v=\frac{\lambda}{T}$ $\mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{sec}$ Distance travelled by wave in time $\mathrm{t}$ is $\mathrm{S}=\mathrm{vt}$ $\mathrm{S}=2 \times 5 \quad$ (Given, $\mathrm{t}=5 \mathrm{sec}$ ) $\mathrm{S}=10 \mathrm{~m}$
JIPMER-2012
WAVES
172298
The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40$ $\pi \mathrm{t}$, where, $\mathrm{x}$ and $\mathrm{y}$ are in $\mathrm{cm}$ and $\mathrm{t}$ in second. The separation between two adjacent nodes is:
1 $1.5 \mathrm{~cm}$
2 $3 \mathrm{~cm}$
3 $6 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Distance between adjacent nodes is half of wavelength. The standard equation of stationary wave is $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{x}}{\lambda} \times \cos \frac{2 \pi \mathrm{t}}{\lambda}$ Where a is amplitude, $\lambda$ is wavelength. Given equation is $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$ Comparing eq $^{\mathrm{n}}$ (i) and eq ${ }^{\mathrm{n}}$ (ii), we get, $\frac{2 \pi}{\lambda}=\frac{\pi}{3}$ $\therefore \quad \lambda=6 \mathrm{~cm}$ Distance between two adjacent nodes $\Rightarrow \frac{\lambda}{2}=\frac{6}{2}$ $\lambda=3 \mathrm{~cm}$
172292
The wavelength of a wave in a medium is $0.5 \mathrm{~m}$. The phase difference between the oscillations at two points in the medium due to this wave is $\frac{\pi}{5}$. What is the minimum distance between these points?
1 $0.05 \mathrm{~m}$
2 $0.1 \mathrm{~m}$
3 $0.25 \mathrm{~m}$
4 $0.15 \mathrm{~m}$
Explanation:
A Given that Wavelength $\lambda=0.5$ Phase change $\Delta \phi=\pi / 5$ $\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ From equation (i) $\pi / 5=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $\Delta \mathrm{x}=\frac{\lambda}{10}$ $=\frac{0.5}{10}=0.05 \mathrm{~m}$
MP PMT-2013
WAVES
172295
If $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right) y \rightarrow m m, t \rightarrow$ second, $\mathbf{x} \rightarrow \mathbf{m}$. For given progressive wave equation, phase difference between two vibrating particle having path difference $3.5 \mathrm{~m}$ would be
1 $\pi / 4$
2 $\pi$
3 $\pi / 3$
4 $\pi / 2$
Explanation:
D Given that $y=5 \sin \left(30 \pi t-\frac{\pi}{7} x+30^{\circ}\right)$ $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Comparing the equation (i) and (ii) $\mathrm{A}=5$ $\omega=30 \pi$ $\mathrm{k}=\pi / 7$ Wave length, $\lambda=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \pi \times 7}{\pi}$ $\lambda=14 \mathrm{~m}$ Relation between phase difference and path difference $\Delta \phi =\frac{2 \pi}{\lambda} \times \Delta x$ $=\frac{2 \pi}{14} \times 3.5$ $\Delta \phi =\frac{\pi}{2}$
MP PET-2008
WAVES
172297
$Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ represents an equation of a progressive wave, where $t$ is in second and $x$ is in metre. The distance travelled by the wave in $5 \mathrm{~s}$ is
1 $8 \mathrm{~m}$
2 $10 \mathrm{~m}$
3 $5 \mathrm{~m}$
4 $32 \mathrm{~m}$
Explanation:
B The given equation of a progressive wave is $Y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$ $\mathrm{Y}=3 \sin 2 \pi\left(\frac{\mathrm{t}}{4}-\frac{\mathrm{x}}{8}\right)$ The standard equation of a progressive wave is $\mathrm{Y}=\mathrm{a} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$ Comparing eq. (i) and (ii) $\mathrm{T}=4 \mathrm{~s} \quad \lambda=8 \mathrm{~m}$ Velocity of wave, $v=\frac{\lambda}{T}$ $\mathrm{v}=\frac{8}{4}=2 \mathrm{~m} / \mathrm{sec}$ Distance travelled by wave in time $\mathrm{t}$ is $\mathrm{S}=\mathrm{vt}$ $\mathrm{S}=2 \times 5 \quad$ (Given, $\mathrm{t}=5 \mathrm{sec}$ ) $\mathrm{S}=10 \mathrm{~m}$
JIPMER-2012
WAVES
172298
The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40$ $\pi \mathrm{t}$, where, $\mathrm{x}$ and $\mathrm{y}$ are in $\mathrm{cm}$ and $\mathrm{t}$ in second. The separation between two adjacent nodes is:
1 $1.5 \mathrm{~cm}$
2 $3 \mathrm{~cm}$
3 $6 \mathrm{~cm}$
4 $4 \mathrm{~cm}$
Explanation:
B Distance between adjacent nodes is half of wavelength. The standard equation of stationary wave is $\mathrm{y}=2 \mathrm{~A} \sin \frac{2 \pi \mathrm{x}}{\lambda} \times \cos \frac{2 \pi \mathrm{t}}{\lambda}$ Where a is amplitude, $\lambda$ is wavelength. Given equation is $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$ Comparing eq $^{\mathrm{n}}$ (i) and eq ${ }^{\mathrm{n}}$ (ii), we get, $\frac{2 \pi}{\lambda}=\frac{\pi}{3}$ $\therefore \quad \lambda=6 \mathrm{~cm}$ Distance between two adjacent nodes $\Rightarrow \frac{\lambda}{2}=\frac{6}{2}$ $\lambda=3 \mathrm{~cm}$
