172318
A transverse wave is represented by the equation $y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?
1 $\lambda=2 \pi y_{0}$
2 $\lambda=\frac{\pi y_{0}}{3}$
3 $\lambda=\frac{\pi y_{0}}{2}$
4 $\lambda=\pi y_{0}$
Explanation:
D $: y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ On comparing the given equation with standard equation $y=a \sin \frac{2 \pi}{\lambda}(v t-x)$ speed of wave $(\mathrm{v})$ wave $=\mathrm{v}$ maximum velocity $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \mathrm{co}-$ efficient of $\mathrm{t}$ $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \frac{2 \pi \mathrm{v}}{\lambda}$ $\left(\mathrm{V}_{\max }\right)_{\mathrm{p}}=2(\mathrm{v})$ wave $\frac{\mathrm{y}_{0} \times 2 \pi \mathrm{v}}{\lambda}=2 \mathrm{v}$ $\lambda=\pi y_{\mathrm{o}}$
AIPMT-1995
WAVES
172319
The equation of a sound wave is given as $y=0.005 \sin (62.4 x+316 t)$. The wavelength of this wave is
1 0.4unit
2 0.3 unit
3 0.2unit
4 0.1unit
Explanation:
D Given equation, $y=0.005 \sin (62.4 x+316 t)$ $y=A \sin (\omega t+k x)$ On comparing, $\omega=316$ $k=62.4$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \times 3.14}{62.4}$ $\lambda =0.1 \text { unit }$
AIPMT-1996
WAVES
172320
The phase difference between two waves, represented by $y_{1}=10^{-6} \sin \left\{100 t+\left(\frac{x}{50}\right)+0.5\right\} m$ $y_{2}=10^{-6} \cos \left\{100 t+\left(\frac{x}{50}\right)\right\} m,$ Where, $x$ is expressed in metre and $t$ is expressed in second, is approximately
1 $1.07 \mathrm{rad}$
2 $2.07 \mathrm{rad}$
3 $0.5 \mathrm{rad}$
4 $1.5 \mathrm{rad}$
Explanation:
A $\mathrm{Y}_{1}=10^{-6} \sin \{100 \mathrm{t}+(\mathrm{x} / 50)+0.5\} \mathrm{m}$.....(i) $\mathrm{Y}_{2}=10^{-6} \cos \{100 \mathrm{t}+(\mathrm{x} / 50)\} \mathrm{m}$ From equation (ii) $Y_{2}=10^{-6} \sin \left[100 t+\frac{x}{50}+\frac{\pi}{2}\right] \mathrm{m}$ From equation (i) $\phi_{1}=0.5 \mathrm{~m}$ from equation (i) $\phi_{2}=\frac{\pi}{2}$ Phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}$ $=\frac{\pi}{2}-0.5$ $\Delta \phi=1.07 \mathrm{rad}$
AIPMT-2004
WAVES
172321
A wave of amplitude $\mathrm{a}=\mathbf{0 . 2 \mathrm { m }}$, velocity $\mathrm{v}=360$ $\mathrm{m} / \mathrm{s}$ and $\lambda$ wavelength $60 \mathrm{~m}$ is travelling along positive $x$-axis, then the correct expression for the wave is
172318
A transverse wave is represented by the equation $y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?
1 $\lambda=2 \pi y_{0}$
2 $\lambda=\frac{\pi y_{0}}{3}$
3 $\lambda=\frac{\pi y_{0}}{2}$
4 $\lambda=\pi y_{0}$
Explanation:
D $: y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ On comparing the given equation with standard equation $y=a \sin \frac{2 \pi}{\lambda}(v t-x)$ speed of wave $(\mathrm{v})$ wave $=\mathrm{v}$ maximum velocity $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \mathrm{co}-$ efficient of $\mathrm{t}$ $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \frac{2 \pi \mathrm{v}}{\lambda}$ $\left(\mathrm{V}_{\max }\right)_{\mathrm{p}}=2(\mathrm{v})$ wave $\frac{\mathrm{y}_{0} \times 2 \pi \mathrm{v}}{\lambda}=2 \mathrm{v}$ $\lambda=\pi y_{\mathrm{o}}$
AIPMT-1995
WAVES
172319
The equation of a sound wave is given as $y=0.005 \sin (62.4 x+316 t)$. The wavelength of this wave is
1 0.4unit
2 0.3 unit
3 0.2unit
4 0.1unit
Explanation:
D Given equation, $y=0.005 \sin (62.4 x+316 t)$ $y=A \sin (\omega t+k x)$ On comparing, $\omega=316$ $k=62.4$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \times 3.14}{62.4}$ $\lambda =0.1 \text { unit }$
AIPMT-1996
WAVES
172320
The phase difference between two waves, represented by $y_{1}=10^{-6} \sin \left\{100 t+\left(\frac{x}{50}\right)+0.5\right\} m$ $y_{2}=10^{-6} \cos \left\{100 t+\left(\frac{x}{50}\right)\right\} m,$ Where, $x$ is expressed in metre and $t$ is expressed in second, is approximately
1 $1.07 \mathrm{rad}$
2 $2.07 \mathrm{rad}$
3 $0.5 \mathrm{rad}$
4 $1.5 \mathrm{rad}$
Explanation:
A $\mathrm{Y}_{1}=10^{-6} \sin \{100 \mathrm{t}+(\mathrm{x} / 50)+0.5\} \mathrm{m}$.....(i) $\mathrm{Y}_{2}=10^{-6} \cos \{100 \mathrm{t}+(\mathrm{x} / 50)\} \mathrm{m}$ From equation (ii) $Y_{2}=10^{-6} \sin \left[100 t+\frac{x}{50}+\frac{\pi}{2}\right] \mathrm{m}$ From equation (i) $\phi_{1}=0.5 \mathrm{~m}$ from equation (i) $\phi_{2}=\frac{\pi}{2}$ Phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}$ $=\frac{\pi}{2}-0.5$ $\Delta \phi=1.07 \mathrm{rad}$
AIPMT-2004
WAVES
172321
A wave of amplitude $\mathrm{a}=\mathbf{0 . 2 \mathrm { m }}$, velocity $\mathrm{v}=360$ $\mathrm{m} / \mathrm{s}$ and $\lambda$ wavelength $60 \mathrm{~m}$ is travelling along positive $x$-axis, then the correct expression for the wave is
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WAVES
172318
A transverse wave is represented by the equation $y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?
1 $\lambda=2 \pi y_{0}$
2 $\lambda=\frac{\pi y_{0}}{3}$
3 $\lambda=\frac{\pi y_{0}}{2}$
4 $\lambda=\pi y_{0}$
Explanation:
D $: y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ On comparing the given equation with standard equation $y=a \sin \frac{2 \pi}{\lambda}(v t-x)$ speed of wave $(\mathrm{v})$ wave $=\mathrm{v}$ maximum velocity $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \mathrm{co}-$ efficient of $\mathrm{t}$ $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \frac{2 \pi \mathrm{v}}{\lambda}$ $\left(\mathrm{V}_{\max }\right)_{\mathrm{p}}=2(\mathrm{v})$ wave $\frac{\mathrm{y}_{0} \times 2 \pi \mathrm{v}}{\lambda}=2 \mathrm{v}$ $\lambda=\pi y_{\mathrm{o}}$
AIPMT-1995
WAVES
172319
The equation of a sound wave is given as $y=0.005 \sin (62.4 x+316 t)$. The wavelength of this wave is
1 0.4unit
2 0.3 unit
3 0.2unit
4 0.1unit
Explanation:
D Given equation, $y=0.005 \sin (62.4 x+316 t)$ $y=A \sin (\omega t+k x)$ On comparing, $\omega=316$ $k=62.4$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \times 3.14}{62.4}$ $\lambda =0.1 \text { unit }$
AIPMT-1996
WAVES
172320
The phase difference between two waves, represented by $y_{1}=10^{-6} \sin \left\{100 t+\left(\frac{x}{50}\right)+0.5\right\} m$ $y_{2}=10^{-6} \cos \left\{100 t+\left(\frac{x}{50}\right)\right\} m,$ Where, $x$ is expressed in metre and $t$ is expressed in second, is approximately
1 $1.07 \mathrm{rad}$
2 $2.07 \mathrm{rad}$
3 $0.5 \mathrm{rad}$
4 $1.5 \mathrm{rad}$
Explanation:
A $\mathrm{Y}_{1}=10^{-6} \sin \{100 \mathrm{t}+(\mathrm{x} / 50)+0.5\} \mathrm{m}$.....(i) $\mathrm{Y}_{2}=10^{-6} \cos \{100 \mathrm{t}+(\mathrm{x} / 50)\} \mathrm{m}$ From equation (ii) $Y_{2}=10^{-6} \sin \left[100 t+\frac{x}{50}+\frac{\pi}{2}\right] \mathrm{m}$ From equation (i) $\phi_{1}=0.5 \mathrm{~m}$ from equation (i) $\phi_{2}=\frac{\pi}{2}$ Phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}$ $=\frac{\pi}{2}-0.5$ $\Delta \phi=1.07 \mathrm{rad}$
AIPMT-2004
WAVES
172321
A wave of amplitude $\mathrm{a}=\mathbf{0 . 2 \mathrm { m }}$, velocity $\mathrm{v}=360$ $\mathrm{m} / \mathrm{s}$ and $\lambda$ wavelength $60 \mathrm{~m}$ is travelling along positive $x$-axis, then the correct expression for the wave is
172318
A transverse wave is represented by the equation $y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?
1 $\lambda=2 \pi y_{0}$
2 $\lambda=\frac{\pi y_{0}}{3}$
3 $\lambda=\frac{\pi y_{0}}{2}$
4 $\lambda=\pi y_{0}$
Explanation:
D $: y=y_{0} \sin \frac{2 \pi}{\lambda}(v t-x)$ On comparing the given equation with standard equation $y=a \sin \frac{2 \pi}{\lambda}(v t-x)$ speed of wave $(\mathrm{v})$ wave $=\mathrm{v}$ maximum velocity $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \mathrm{co}-$ efficient of $\mathrm{t}$ $\mathrm{V}_{\max }=\mathrm{a} \omega=\mathrm{y}_{\mathrm{o}} \times \frac{2 \pi \mathrm{v}}{\lambda}$ $\left(\mathrm{V}_{\max }\right)_{\mathrm{p}}=2(\mathrm{v})$ wave $\frac{\mathrm{y}_{0} \times 2 \pi \mathrm{v}}{\lambda}=2 \mathrm{v}$ $\lambda=\pi y_{\mathrm{o}}$
AIPMT-1995
WAVES
172319
The equation of a sound wave is given as $y=0.005 \sin (62.4 x+316 t)$. The wavelength of this wave is
1 0.4unit
2 0.3 unit
3 0.2unit
4 0.1unit
Explanation:
D Given equation, $y=0.005 \sin (62.4 x+316 t)$ $y=A \sin (\omega t+k x)$ On comparing, $\omega=316$ $k=62.4$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}$ $=\frac{2 \times 3.14}{62.4}$ $\lambda =0.1 \text { unit }$
AIPMT-1996
WAVES
172320
The phase difference between two waves, represented by $y_{1}=10^{-6} \sin \left\{100 t+\left(\frac{x}{50}\right)+0.5\right\} m$ $y_{2}=10^{-6} \cos \left\{100 t+\left(\frac{x}{50}\right)\right\} m,$ Where, $x$ is expressed in metre and $t$ is expressed in second, is approximately
1 $1.07 \mathrm{rad}$
2 $2.07 \mathrm{rad}$
3 $0.5 \mathrm{rad}$
4 $1.5 \mathrm{rad}$
Explanation:
A $\mathrm{Y}_{1}=10^{-6} \sin \{100 \mathrm{t}+(\mathrm{x} / 50)+0.5\} \mathrm{m}$.....(i) $\mathrm{Y}_{2}=10^{-6} \cos \{100 \mathrm{t}+(\mathrm{x} / 50)\} \mathrm{m}$ From equation (ii) $Y_{2}=10^{-6} \sin \left[100 t+\frac{x}{50}+\frac{\pi}{2}\right] \mathrm{m}$ From equation (i) $\phi_{1}=0.5 \mathrm{~m}$ from equation (i) $\phi_{2}=\frac{\pi}{2}$ Phase difference $(\Delta \phi)=\phi_{2}-\phi_{1}$ $=\frac{\pi}{2}-0.5$ $\Delta \phi=1.07 \mathrm{rad}$
AIPMT-2004
WAVES
172321
A wave of amplitude $\mathrm{a}=\mathbf{0 . 2 \mathrm { m }}$, velocity $\mathrm{v}=360$ $\mathrm{m} / \mathrm{s}$ and $\lambda$ wavelength $60 \mathrm{~m}$ is travelling along positive $x$-axis, then the correct expression for the wave is