172322
The equation of a wave is given by $y=a$ sin $\left(100 t-\frac{x}{10}\right)$, where $x$ and $y$ are in metre and $t$ in second, then velocity of wave is
1 $0.1 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $100 \mathrm{~m} / \mathrm{s}$
4 $1000 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given equation, $y=a \sin \left(100 t-\frac{x}{10}\right)$ Standard equation of the wave $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=100, \mathrm{k}=\frac{1}{10}$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{\frac{1}{10}}=1000$ $\mathrm{v}=1000 \mathrm{~m} / \mathrm{sec}$
AIPMT-2001
WAVES
172324
A transverse wave propagating along $x$-axis is represented by $y(x, t)=8 \sin \left(0.5 \pi x-4 \pi t-\frac{\pi}{4}\right)$ Where, $x$ is in metre and $t$ is in second. The speed the wave is
1 $4 \pi \mathrm{m} / \mathrm{s}$
2 $0.5 \pi \mathrm{m} / \mathrm{s}$
3 $\frac{\pi}{4} \mathrm{~m} / \mathrm{s}$
4 $8 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Equation of transverse wave along $x$-axis $y(x, t)=8 \sin (0.5 \pi x-4 \pi t-\pi / 4)$ Standard equation of the wave $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ $\mathrm{k}=0.5 \pi, \omega=4 \pi$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{4 \pi}{0.5 \pi}$ $\mathrm{v}=\frac{4}{0.5}$ $\mathrm{v}=8 \mathrm{~m} / \text { second. }$
AIPMT-2006
WAVES
172325
Two waves are represented by the equations $Y_{1}$ $=\mathbf{a} \sin (\omega t+\mathbf{k x}+0.57) \mathrm{m}$ and $Y_{2}=\mathbf{a} \cos (\omega t+$ kx) $m$, where $x$ is in metre and $t$ in second. The phase difference between them is
1 $1.26 \mathrm{rad}$
2 $1.57 \mathrm{rad}$
3 $0.57 \mathrm{rad}$
4 $1 \mathrm{rad}$
Explanation:
D Given Equation, $Y_{1}=a \sin (\omega t+k x+0.57) \ldots(i)$ $Y_{2}=a \cos (\omega t+k x)$ $Y_{2}=a \sin (\omega t+k x+\pi / 2)$ From equation (i) and (ii), $\phi_{1}=0.57$ $\phi_{2}=\pi / 2$ Then phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}=\pi / 2-0.57=1$ radian
AIPMT-2011
WAVES
172326
Sound waves travel at $350 \mathrm{~m} / \mathrm{s}$ through a warm air and at $3500 \mathrm{~m} / \mathrm{s}$ through brass. The wavelength of a $700 \mathrm{~Hz}$ acoustic wave as it enters brass from warm air
1 increases by a factor 20
2 increases by a factor 10
3 decreases by a factor 20
4 decreases by a factor 10
Explanation:
B Given that, $\mathrm{v}_{1}=350 \mathrm{~m} / \mathrm{sec}, \mathrm{v}_{2}=3500 \mathrm{~m} / \mathrm{sec}$ Sound wave enter brass from warm air frequency of the medium does not change We know that, $\mathrm{v}=f \lambda$ For first medium, $\mathrm{v}_{1}=f \lambda_{1}$ For second medium, $\left(\mathrm{v}_{2}\right)=f \lambda_{2}$ Then, $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} =\frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2} =\lambda_{1} \times \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ $=\lambda_{1} \times \frac{3500}{350}$ $\lambda_{2} =10 \lambda_{1}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
WAVES
172322
The equation of a wave is given by $y=a$ sin $\left(100 t-\frac{x}{10}\right)$, where $x$ and $y$ are in metre and $t$ in second, then velocity of wave is
1 $0.1 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $100 \mathrm{~m} / \mathrm{s}$
4 $1000 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given equation, $y=a \sin \left(100 t-\frac{x}{10}\right)$ Standard equation of the wave $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=100, \mathrm{k}=\frac{1}{10}$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{\frac{1}{10}}=1000$ $\mathrm{v}=1000 \mathrm{~m} / \mathrm{sec}$
AIPMT-2001
WAVES
172324
A transverse wave propagating along $x$-axis is represented by $y(x, t)=8 \sin \left(0.5 \pi x-4 \pi t-\frac{\pi}{4}\right)$ Where, $x$ is in metre and $t$ is in second. The speed the wave is
1 $4 \pi \mathrm{m} / \mathrm{s}$
2 $0.5 \pi \mathrm{m} / \mathrm{s}$
3 $\frac{\pi}{4} \mathrm{~m} / \mathrm{s}$
4 $8 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Equation of transverse wave along $x$-axis $y(x, t)=8 \sin (0.5 \pi x-4 \pi t-\pi / 4)$ Standard equation of the wave $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ $\mathrm{k}=0.5 \pi, \omega=4 \pi$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{4 \pi}{0.5 \pi}$ $\mathrm{v}=\frac{4}{0.5}$ $\mathrm{v}=8 \mathrm{~m} / \text { second. }$
AIPMT-2006
WAVES
172325
Two waves are represented by the equations $Y_{1}$ $=\mathbf{a} \sin (\omega t+\mathbf{k x}+0.57) \mathrm{m}$ and $Y_{2}=\mathbf{a} \cos (\omega t+$ kx) $m$, where $x$ is in metre and $t$ in second. The phase difference between them is
1 $1.26 \mathrm{rad}$
2 $1.57 \mathrm{rad}$
3 $0.57 \mathrm{rad}$
4 $1 \mathrm{rad}$
Explanation:
D Given Equation, $Y_{1}=a \sin (\omega t+k x+0.57) \ldots(i)$ $Y_{2}=a \cos (\omega t+k x)$ $Y_{2}=a \sin (\omega t+k x+\pi / 2)$ From equation (i) and (ii), $\phi_{1}=0.57$ $\phi_{2}=\pi / 2$ Then phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}=\pi / 2-0.57=1$ radian
AIPMT-2011
WAVES
172326
Sound waves travel at $350 \mathrm{~m} / \mathrm{s}$ through a warm air and at $3500 \mathrm{~m} / \mathrm{s}$ through brass. The wavelength of a $700 \mathrm{~Hz}$ acoustic wave as it enters brass from warm air
1 increases by a factor 20
2 increases by a factor 10
3 decreases by a factor 20
4 decreases by a factor 10
Explanation:
B Given that, $\mathrm{v}_{1}=350 \mathrm{~m} / \mathrm{sec}, \mathrm{v}_{2}=3500 \mathrm{~m} / \mathrm{sec}$ Sound wave enter brass from warm air frequency of the medium does not change We know that, $\mathrm{v}=f \lambda$ For first medium, $\mathrm{v}_{1}=f \lambda_{1}$ For second medium, $\left(\mathrm{v}_{2}\right)=f \lambda_{2}$ Then, $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} =\frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2} =\lambda_{1} \times \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ $=\lambda_{1} \times \frac{3500}{350}$ $\lambda_{2} =10 \lambda_{1}$
172322
The equation of a wave is given by $y=a$ sin $\left(100 t-\frac{x}{10}\right)$, where $x$ and $y$ are in metre and $t$ in second, then velocity of wave is
1 $0.1 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $100 \mathrm{~m} / \mathrm{s}$
4 $1000 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given equation, $y=a \sin \left(100 t-\frac{x}{10}\right)$ Standard equation of the wave $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=100, \mathrm{k}=\frac{1}{10}$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{\frac{1}{10}}=1000$ $\mathrm{v}=1000 \mathrm{~m} / \mathrm{sec}$
AIPMT-2001
WAVES
172324
A transverse wave propagating along $x$-axis is represented by $y(x, t)=8 \sin \left(0.5 \pi x-4 \pi t-\frac{\pi}{4}\right)$ Where, $x$ is in metre and $t$ is in second. The speed the wave is
1 $4 \pi \mathrm{m} / \mathrm{s}$
2 $0.5 \pi \mathrm{m} / \mathrm{s}$
3 $\frac{\pi}{4} \mathrm{~m} / \mathrm{s}$
4 $8 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Equation of transverse wave along $x$-axis $y(x, t)=8 \sin (0.5 \pi x-4 \pi t-\pi / 4)$ Standard equation of the wave $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ $\mathrm{k}=0.5 \pi, \omega=4 \pi$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{4 \pi}{0.5 \pi}$ $\mathrm{v}=\frac{4}{0.5}$ $\mathrm{v}=8 \mathrm{~m} / \text { second. }$
AIPMT-2006
WAVES
172325
Two waves are represented by the equations $Y_{1}$ $=\mathbf{a} \sin (\omega t+\mathbf{k x}+0.57) \mathrm{m}$ and $Y_{2}=\mathbf{a} \cos (\omega t+$ kx) $m$, where $x$ is in metre and $t$ in second. The phase difference between them is
1 $1.26 \mathrm{rad}$
2 $1.57 \mathrm{rad}$
3 $0.57 \mathrm{rad}$
4 $1 \mathrm{rad}$
Explanation:
D Given Equation, $Y_{1}=a \sin (\omega t+k x+0.57) \ldots(i)$ $Y_{2}=a \cos (\omega t+k x)$ $Y_{2}=a \sin (\omega t+k x+\pi / 2)$ From equation (i) and (ii), $\phi_{1}=0.57$ $\phi_{2}=\pi / 2$ Then phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}=\pi / 2-0.57=1$ radian
AIPMT-2011
WAVES
172326
Sound waves travel at $350 \mathrm{~m} / \mathrm{s}$ through a warm air and at $3500 \mathrm{~m} / \mathrm{s}$ through brass. The wavelength of a $700 \mathrm{~Hz}$ acoustic wave as it enters brass from warm air
1 increases by a factor 20
2 increases by a factor 10
3 decreases by a factor 20
4 decreases by a factor 10
Explanation:
B Given that, $\mathrm{v}_{1}=350 \mathrm{~m} / \mathrm{sec}, \mathrm{v}_{2}=3500 \mathrm{~m} / \mathrm{sec}$ Sound wave enter brass from warm air frequency of the medium does not change We know that, $\mathrm{v}=f \lambda$ For first medium, $\mathrm{v}_{1}=f \lambda_{1}$ For second medium, $\left(\mathrm{v}_{2}\right)=f \lambda_{2}$ Then, $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} =\frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2} =\lambda_{1} \times \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ $=\lambda_{1} \times \frac{3500}{350}$ $\lambda_{2} =10 \lambda_{1}$
172322
The equation of a wave is given by $y=a$ sin $\left(100 t-\frac{x}{10}\right)$, where $x$ and $y$ are in metre and $t$ in second, then velocity of wave is
1 $0.1 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $100 \mathrm{~m} / \mathrm{s}$
4 $1000 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given equation, $y=a \sin \left(100 t-\frac{x}{10}\right)$ Standard equation of the wave $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\omega=100, \mathrm{k}=\frac{1}{10}$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{100}{\frac{1}{10}}=1000$ $\mathrm{v}=1000 \mathrm{~m} / \mathrm{sec}$
AIPMT-2001
WAVES
172324
A transverse wave propagating along $x$-axis is represented by $y(x, t)=8 \sin \left(0.5 \pi x-4 \pi t-\frac{\pi}{4}\right)$ Where, $x$ is in metre and $t$ is in second. The speed the wave is
1 $4 \pi \mathrm{m} / \mathrm{s}$
2 $0.5 \pi \mathrm{m} / \mathrm{s}$
3 $\frac{\pi}{4} \mathrm{~m} / \mathrm{s}$
4 $8 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Equation of transverse wave along $x$-axis $y(x, t)=8 \sin (0.5 \pi x-4 \pi t-\pi / 4)$ Standard equation of the wave $\mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}+\phi)$ $\mathrm{k}=0.5 \pi, \omega=4 \pi$ $\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{4 \pi}{0.5 \pi}$ $\mathrm{v}=\frac{4}{0.5}$ $\mathrm{v}=8 \mathrm{~m} / \text { second. }$
AIPMT-2006
WAVES
172325
Two waves are represented by the equations $Y_{1}$ $=\mathbf{a} \sin (\omega t+\mathbf{k x}+0.57) \mathrm{m}$ and $Y_{2}=\mathbf{a} \cos (\omega t+$ kx) $m$, where $x$ is in metre and $t$ in second. The phase difference between them is
1 $1.26 \mathrm{rad}$
2 $1.57 \mathrm{rad}$
3 $0.57 \mathrm{rad}$
4 $1 \mathrm{rad}$
Explanation:
D Given Equation, $Y_{1}=a \sin (\omega t+k x+0.57) \ldots(i)$ $Y_{2}=a \cos (\omega t+k x)$ $Y_{2}=a \sin (\omega t+k x+\pi / 2)$ From equation (i) and (ii), $\phi_{1}=0.57$ $\phi_{2}=\pi / 2$ Then phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}=\pi / 2-0.57=1$ radian
AIPMT-2011
WAVES
172326
Sound waves travel at $350 \mathrm{~m} / \mathrm{s}$ through a warm air and at $3500 \mathrm{~m} / \mathrm{s}$ through brass. The wavelength of a $700 \mathrm{~Hz}$ acoustic wave as it enters brass from warm air
1 increases by a factor 20
2 increases by a factor 10
3 decreases by a factor 20
4 decreases by a factor 10
Explanation:
B Given that, $\mathrm{v}_{1}=350 \mathrm{~m} / \mathrm{sec}, \mathrm{v}_{2}=3500 \mathrm{~m} / \mathrm{sec}$ Sound wave enter brass from warm air frequency of the medium does not change We know that, $\mathrm{v}=f \lambda$ For first medium, $\mathrm{v}_{1}=f \lambda_{1}$ For second medium, $\left(\mathrm{v}_{2}\right)=f \lambda_{2}$ Then, $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} =\frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2} =\lambda_{1} \times \frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ $=\lambda_{1} \times \frac{3500}{350}$ $\lambda_{2} =10 \lambda_{1}$
