172182
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is
1 $330 \mathrm{~Hz}$
2 $342 \mathrm{~Hz}$
3 $365 \mathrm{~Hz}$
4 $660 \mathrm{~Hz}$
Explanation:
A The given progressive wave equation, $y=15 \sin (660 \pi t-0.02 \pi x)$ Wave equation, $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$ Comparing equation (ii) with (i) $\omega=660 \pi$ $\frac{2 \pi}{\mathrm{T}}=660 \pi$ $\frac{2}{\mathrm{~T}}=660$ $\frac{1}{\mathrm{~T}}=330 \quad(\therefore 1 / \mathrm{T}=\mathrm{f})$ Frequency $(\mathrm{f})=330 \mathrm{~Hz}$
COMEDK 2020
WAVES
172183
The transverse displacement $y(x, t)$ a wave on a string is given by $y(x, t)=\mathrm{e}^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$. This represents a
1 Wave moving in negative $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
2 Standing wave of frequency $\sqrt{b}$
3 Standing wave of frequency $\frac{1}{\sqrt{b}}$
4 Wave moving in positive $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Explanation:
A Given, $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$ $y(x, t)=e^{\left.-\left\{(\sqrt{a} x)^{2}\right)+(\sqrt{b} t)^{2}+2 \sqrt{a b} x t\right\}}$ $y(x, t)=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ $\text { Speed }=\frac{\text { Coeffcient of }(t)}{\text { Coeffcient of }(x)}$ $\text { Speed }=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$ Hence, option (a) is correct.
TS- EAMCET-10.09.2020
WAVES
172185
Two sinusoidal waves of equal amplitude $Y_{m}$ and wavelength $\lambda$ travel in opposite direction along a stretched string to produce standing waves. The third antinode is located at a distance from one end is
1 $3 / 2 \lambda$
2 $5 / 4 \lambda$
3 $7 / 2 \lambda$
4 $3 \lambda$
Explanation:
B Form the figure Distance between third antinodes $l_{3}=\frac{\lambda}{2}+\frac{\lambda}{4}+\frac{\lambda}{2}$ $l_{3}=\frac{2 \lambda+\lambda+2 \lambda}{4}$ $l_{3}=\frac{5 \lambda}{4}$
AMU-2019
WAVES
172186
The equation of a stationary wave is $y=2 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) . \quad$ The distance between a node and its next antinode is :
1 7.5 units
2 1.5 units
3 22.5 units
4 30 units
Explanation:
A $\mathrm{y}=2 \sin \left(\frac{\pi \mathrm{x}}{15}\right) \cos (48 \pi \mathrm{t})$ Comparing the given equation with standard stationary wave equation we get $\frac{2 \pi}{\lambda}=\frac{\pi}{15}$ $\lambda=30$ Node and antinode distance. $\frac{\lambda}{4}=\frac{30}{4}=7.5 \text { unit }$
Karnataka CET-2019
WAVES
172188
A wave along a string has the following equation $y=0.05 \sin (28 t-2.0 x) m$ (where $t$ is in seconds and $x$ is in meters). What are the amplitude, frequency and wavelength of the wave?
1 amplitude $=0.05 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
2 amplitude $=0.05 \mathrm{~m}$, frequency $=28 \mathrm{~Hz}$ and wavelength $=2.0 \mathrm{~m}$
3 amplitude $=5.0 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
4 amplitude $=0.05 \mathrm{~m}$, frequency $=2.0 \mathrm{~Hz}$ and wavelength $=28 \mathrm{~m}$
5 amplitude $=0.05 \mathrm{~m}$, frequency $=3.456 \mathrm{~Hz}$ and wavelength $=4.518 \mathrm{~m}$
Explanation:
A Given, The equation of wave along string $\mathrm{y}=0.05 \sin (28 \mathrm{t}-2.0 \mathrm{x})$ The general equation of wave is $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Now, comparing eq ${ }^{\mathrm{n}}$. (i) \& (ii) we get- $\mathrm{A}=0.05 \mathrm{~m}, \omega=28, \mathrm{k}=2$ Amplitude $(\mathrm{A})=0.05 \mathrm{~m}$ Frequency $(\mathrm{f})=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{2}=3.14 \mathrm{~m} \square 3.518 \mathrm{~m}$
172182
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is
1 $330 \mathrm{~Hz}$
2 $342 \mathrm{~Hz}$
3 $365 \mathrm{~Hz}$
4 $660 \mathrm{~Hz}$
Explanation:
A The given progressive wave equation, $y=15 \sin (660 \pi t-0.02 \pi x)$ Wave equation, $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$ Comparing equation (ii) with (i) $\omega=660 \pi$ $\frac{2 \pi}{\mathrm{T}}=660 \pi$ $\frac{2}{\mathrm{~T}}=660$ $\frac{1}{\mathrm{~T}}=330 \quad(\therefore 1 / \mathrm{T}=\mathrm{f})$ Frequency $(\mathrm{f})=330 \mathrm{~Hz}$
COMEDK 2020
WAVES
172183
The transverse displacement $y(x, t)$ a wave on a string is given by $y(x, t)=\mathrm{e}^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$. This represents a
1 Wave moving in negative $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
2 Standing wave of frequency $\sqrt{b}$
3 Standing wave of frequency $\frac{1}{\sqrt{b}}$
4 Wave moving in positive $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Explanation:
A Given, $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$ $y(x, t)=e^{\left.-\left\{(\sqrt{a} x)^{2}\right)+(\sqrt{b} t)^{2}+2 \sqrt{a b} x t\right\}}$ $y(x, t)=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ $\text { Speed }=\frac{\text { Coeffcient of }(t)}{\text { Coeffcient of }(x)}$ $\text { Speed }=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$ Hence, option (a) is correct.
TS- EAMCET-10.09.2020
WAVES
172185
Two sinusoidal waves of equal amplitude $Y_{m}$ and wavelength $\lambda$ travel in opposite direction along a stretched string to produce standing waves. The third antinode is located at a distance from one end is
1 $3 / 2 \lambda$
2 $5 / 4 \lambda$
3 $7 / 2 \lambda$
4 $3 \lambda$
Explanation:
B Form the figure Distance between third antinodes $l_{3}=\frac{\lambda}{2}+\frac{\lambda}{4}+\frac{\lambda}{2}$ $l_{3}=\frac{2 \lambda+\lambda+2 \lambda}{4}$ $l_{3}=\frac{5 \lambda}{4}$
AMU-2019
WAVES
172186
The equation of a stationary wave is $y=2 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) . \quad$ The distance between a node and its next antinode is :
1 7.5 units
2 1.5 units
3 22.5 units
4 30 units
Explanation:
A $\mathrm{y}=2 \sin \left(\frac{\pi \mathrm{x}}{15}\right) \cos (48 \pi \mathrm{t})$ Comparing the given equation with standard stationary wave equation we get $\frac{2 \pi}{\lambda}=\frac{\pi}{15}$ $\lambda=30$ Node and antinode distance. $\frac{\lambda}{4}=\frac{30}{4}=7.5 \text { unit }$
Karnataka CET-2019
WAVES
172188
A wave along a string has the following equation $y=0.05 \sin (28 t-2.0 x) m$ (where $t$ is in seconds and $x$ is in meters). What are the amplitude, frequency and wavelength of the wave?
1 amplitude $=0.05 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
2 amplitude $=0.05 \mathrm{~m}$, frequency $=28 \mathrm{~Hz}$ and wavelength $=2.0 \mathrm{~m}$
3 amplitude $=5.0 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
4 amplitude $=0.05 \mathrm{~m}$, frequency $=2.0 \mathrm{~Hz}$ and wavelength $=28 \mathrm{~m}$
5 amplitude $=0.05 \mathrm{~m}$, frequency $=3.456 \mathrm{~Hz}$ and wavelength $=4.518 \mathrm{~m}$
Explanation:
A Given, The equation of wave along string $\mathrm{y}=0.05 \sin (28 \mathrm{t}-2.0 \mathrm{x})$ The general equation of wave is $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Now, comparing eq ${ }^{\mathrm{n}}$. (i) \& (ii) we get- $\mathrm{A}=0.05 \mathrm{~m}, \omega=28, \mathrm{k}=2$ Amplitude $(\mathrm{A})=0.05 \mathrm{~m}$ Frequency $(\mathrm{f})=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{2}=3.14 \mathrm{~m} \square 3.518 \mathrm{~m}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172182
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is
1 $330 \mathrm{~Hz}$
2 $342 \mathrm{~Hz}$
3 $365 \mathrm{~Hz}$
4 $660 \mathrm{~Hz}$
Explanation:
A The given progressive wave equation, $y=15 \sin (660 \pi t-0.02 \pi x)$ Wave equation, $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$ Comparing equation (ii) with (i) $\omega=660 \pi$ $\frac{2 \pi}{\mathrm{T}}=660 \pi$ $\frac{2}{\mathrm{~T}}=660$ $\frac{1}{\mathrm{~T}}=330 \quad(\therefore 1 / \mathrm{T}=\mathrm{f})$ Frequency $(\mathrm{f})=330 \mathrm{~Hz}$
COMEDK 2020
WAVES
172183
The transverse displacement $y(x, t)$ a wave on a string is given by $y(x, t)=\mathrm{e}^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$. This represents a
1 Wave moving in negative $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
2 Standing wave of frequency $\sqrt{b}$
3 Standing wave of frequency $\frac{1}{\sqrt{b}}$
4 Wave moving in positive $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Explanation:
A Given, $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$ $y(x, t)=e^{\left.-\left\{(\sqrt{a} x)^{2}\right)+(\sqrt{b} t)^{2}+2 \sqrt{a b} x t\right\}}$ $y(x, t)=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ $\text { Speed }=\frac{\text { Coeffcient of }(t)}{\text { Coeffcient of }(x)}$ $\text { Speed }=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$ Hence, option (a) is correct.
TS- EAMCET-10.09.2020
WAVES
172185
Two sinusoidal waves of equal amplitude $Y_{m}$ and wavelength $\lambda$ travel in opposite direction along a stretched string to produce standing waves. The third antinode is located at a distance from one end is
1 $3 / 2 \lambda$
2 $5 / 4 \lambda$
3 $7 / 2 \lambda$
4 $3 \lambda$
Explanation:
B Form the figure Distance between third antinodes $l_{3}=\frac{\lambda}{2}+\frac{\lambda}{4}+\frac{\lambda}{2}$ $l_{3}=\frac{2 \lambda+\lambda+2 \lambda}{4}$ $l_{3}=\frac{5 \lambda}{4}$
AMU-2019
WAVES
172186
The equation of a stationary wave is $y=2 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) . \quad$ The distance between a node and its next antinode is :
1 7.5 units
2 1.5 units
3 22.5 units
4 30 units
Explanation:
A $\mathrm{y}=2 \sin \left(\frac{\pi \mathrm{x}}{15}\right) \cos (48 \pi \mathrm{t})$ Comparing the given equation with standard stationary wave equation we get $\frac{2 \pi}{\lambda}=\frac{\pi}{15}$ $\lambda=30$ Node and antinode distance. $\frac{\lambda}{4}=\frac{30}{4}=7.5 \text { unit }$
Karnataka CET-2019
WAVES
172188
A wave along a string has the following equation $y=0.05 \sin (28 t-2.0 x) m$ (where $t$ is in seconds and $x$ is in meters). What are the amplitude, frequency and wavelength of the wave?
1 amplitude $=0.05 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
2 amplitude $=0.05 \mathrm{~m}$, frequency $=28 \mathrm{~Hz}$ and wavelength $=2.0 \mathrm{~m}$
3 amplitude $=5.0 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
4 amplitude $=0.05 \mathrm{~m}$, frequency $=2.0 \mathrm{~Hz}$ and wavelength $=28 \mathrm{~m}$
5 amplitude $=0.05 \mathrm{~m}$, frequency $=3.456 \mathrm{~Hz}$ and wavelength $=4.518 \mathrm{~m}$
Explanation:
A Given, The equation of wave along string $\mathrm{y}=0.05 \sin (28 \mathrm{t}-2.0 \mathrm{x})$ The general equation of wave is $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Now, comparing eq ${ }^{\mathrm{n}}$. (i) \& (ii) we get- $\mathrm{A}=0.05 \mathrm{~m}, \omega=28, \mathrm{k}=2$ Amplitude $(\mathrm{A})=0.05 \mathrm{~m}$ Frequency $(\mathrm{f})=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{2}=3.14 \mathrm{~m} \square 3.518 \mathrm{~m}$
172182
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is
1 $330 \mathrm{~Hz}$
2 $342 \mathrm{~Hz}$
3 $365 \mathrm{~Hz}$
4 $660 \mathrm{~Hz}$
Explanation:
A The given progressive wave equation, $y=15 \sin (660 \pi t-0.02 \pi x)$ Wave equation, $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$ Comparing equation (ii) with (i) $\omega=660 \pi$ $\frac{2 \pi}{\mathrm{T}}=660 \pi$ $\frac{2}{\mathrm{~T}}=660$ $\frac{1}{\mathrm{~T}}=330 \quad(\therefore 1 / \mathrm{T}=\mathrm{f})$ Frequency $(\mathrm{f})=330 \mathrm{~Hz}$
COMEDK 2020
WAVES
172183
The transverse displacement $y(x, t)$ a wave on a string is given by $y(x, t)=\mathrm{e}^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$. This represents a
1 Wave moving in negative $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
2 Standing wave of frequency $\sqrt{b}$
3 Standing wave of frequency $\frac{1}{\sqrt{b}}$
4 Wave moving in positive $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Explanation:
A Given, $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$ $y(x, t)=e^{\left.-\left\{(\sqrt{a} x)^{2}\right)+(\sqrt{b} t)^{2}+2 \sqrt{a b} x t\right\}}$ $y(x, t)=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ $\text { Speed }=\frac{\text { Coeffcient of }(t)}{\text { Coeffcient of }(x)}$ $\text { Speed }=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$ Hence, option (a) is correct.
TS- EAMCET-10.09.2020
WAVES
172185
Two sinusoidal waves of equal amplitude $Y_{m}$ and wavelength $\lambda$ travel in opposite direction along a stretched string to produce standing waves. The third antinode is located at a distance from one end is
1 $3 / 2 \lambda$
2 $5 / 4 \lambda$
3 $7 / 2 \lambda$
4 $3 \lambda$
Explanation:
B Form the figure Distance between third antinodes $l_{3}=\frac{\lambda}{2}+\frac{\lambda}{4}+\frac{\lambda}{2}$ $l_{3}=\frac{2 \lambda+\lambda+2 \lambda}{4}$ $l_{3}=\frac{5 \lambda}{4}$
AMU-2019
WAVES
172186
The equation of a stationary wave is $y=2 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) . \quad$ The distance between a node and its next antinode is :
1 7.5 units
2 1.5 units
3 22.5 units
4 30 units
Explanation:
A $\mathrm{y}=2 \sin \left(\frac{\pi \mathrm{x}}{15}\right) \cos (48 \pi \mathrm{t})$ Comparing the given equation with standard stationary wave equation we get $\frac{2 \pi}{\lambda}=\frac{\pi}{15}$ $\lambda=30$ Node and antinode distance. $\frac{\lambda}{4}=\frac{30}{4}=7.5 \text { unit }$
Karnataka CET-2019
WAVES
172188
A wave along a string has the following equation $y=0.05 \sin (28 t-2.0 x) m$ (where $t$ is in seconds and $x$ is in meters). What are the amplitude, frequency and wavelength of the wave?
1 amplitude $=0.05 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
2 amplitude $=0.05 \mathrm{~m}$, frequency $=28 \mathrm{~Hz}$ and wavelength $=2.0 \mathrm{~m}$
3 amplitude $=5.0 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
4 amplitude $=0.05 \mathrm{~m}$, frequency $=2.0 \mathrm{~Hz}$ and wavelength $=28 \mathrm{~m}$
5 amplitude $=0.05 \mathrm{~m}$, frequency $=3.456 \mathrm{~Hz}$ and wavelength $=4.518 \mathrm{~m}$
Explanation:
A Given, The equation of wave along string $\mathrm{y}=0.05 \sin (28 \mathrm{t}-2.0 \mathrm{x})$ The general equation of wave is $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Now, comparing eq ${ }^{\mathrm{n}}$. (i) \& (ii) we get- $\mathrm{A}=0.05 \mathrm{~m}, \omega=28, \mathrm{k}=2$ Amplitude $(\mathrm{A})=0.05 \mathrm{~m}$ Frequency $(\mathrm{f})=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{2}=3.14 \mathrm{~m} \square 3.518 \mathrm{~m}$
172182
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is
1 $330 \mathrm{~Hz}$
2 $342 \mathrm{~Hz}$
3 $365 \mathrm{~Hz}$
4 $660 \mathrm{~Hz}$
Explanation:
A The given progressive wave equation, $y=15 \sin (660 \pi t-0.02 \pi x)$ Wave equation, $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\phi)$ Comparing equation (ii) with (i) $\omega=660 \pi$ $\frac{2 \pi}{\mathrm{T}}=660 \pi$ $\frac{2}{\mathrm{~T}}=660$ $\frac{1}{\mathrm{~T}}=330 \quad(\therefore 1 / \mathrm{T}=\mathrm{f})$ Frequency $(\mathrm{f})=330 \mathrm{~Hz}$
COMEDK 2020
WAVES
172183
The transverse displacement $y(x, t)$ a wave on a string is given by $y(x, t)=\mathrm{e}^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$. This represents a
1 Wave moving in negative $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
2 Standing wave of frequency $\sqrt{b}$
3 Standing wave of frequency $\frac{1}{\sqrt{b}}$
4 Wave moving in positive $\mathrm{x}$-direction with speed $\sqrt{\frac{b}{a}}$
Explanation:
A Given, $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$ $y(x, t)=e^{\left.-\left\{(\sqrt{a} x)^{2}\right)+(\sqrt{b} t)^{2}+2 \sqrt{a b} x t\right\}}$ $y(x, t)=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ $\text { Speed }=\frac{\text { Coeffcient of }(t)}{\text { Coeffcient of }(x)}$ $\text { Speed }=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$ Hence, option (a) is correct.
TS- EAMCET-10.09.2020
WAVES
172185
Two sinusoidal waves of equal amplitude $Y_{m}$ and wavelength $\lambda$ travel in opposite direction along a stretched string to produce standing waves. The third antinode is located at a distance from one end is
1 $3 / 2 \lambda$
2 $5 / 4 \lambda$
3 $7 / 2 \lambda$
4 $3 \lambda$
Explanation:
B Form the figure Distance between third antinodes $l_{3}=\frac{\lambda}{2}+\frac{\lambda}{4}+\frac{\lambda}{2}$ $l_{3}=\frac{2 \lambda+\lambda+2 \lambda}{4}$ $l_{3}=\frac{5 \lambda}{4}$
AMU-2019
WAVES
172186
The equation of a stationary wave is $y=2 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) . \quad$ The distance between a node and its next antinode is :
1 7.5 units
2 1.5 units
3 22.5 units
4 30 units
Explanation:
A $\mathrm{y}=2 \sin \left(\frac{\pi \mathrm{x}}{15}\right) \cos (48 \pi \mathrm{t})$ Comparing the given equation with standard stationary wave equation we get $\frac{2 \pi}{\lambda}=\frac{\pi}{15}$ $\lambda=30$ Node and antinode distance. $\frac{\lambda}{4}=\frac{30}{4}=7.5 \text { unit }$
Karnataka CET-2019
WAVES
172188
A wave along a string has the following equation $y=0.05 \sin (28 t-2.0 x) m$ (where $t$ is in seconds and $x$ is in meters). What are the amplitude, frequency and wavelength of the wave?
1 amplitude $=0.05 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
2 amplitude $=0.05 \mathrm{~m}$, frequency $=28 \mathrm{~Hz}$ and wavelength $=2.0 \mathrm{~m}$
3 amplitude $=5.0 \mathrm{~m}$, frequency $=4.456 \mathrm{~Hz}$ and wavelength $=3.518 \mathrm{~m}$
4 amplitude $=0.05 \mathrm{~m}$, frequency $=2.0 \mathrm{~Hz}$ and wavelength $=28 \mathrm{~m}$
5 amplitude $=0.05 \mathrm{~m}$, frequency $=3.456 \mathrm{~Hz}$ and wavelength $=4.518 \mathrm{~m}$
Explanation:
A Given, The equation of wave along string $\mathrm{y}=0.05 \sin (28 \mathrm{t}-2.0 \mathrm{x})$ The general equation of wave is $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Now, comparing eq ${ }^{\mathrm{n}}$. (i) \& (ii) we get- $\mathrm{A}=0.05 \mathrm{~m}, \omega=28, \mathrm{k}=2$ Amplitude $(\mathrm{A})=0.05 \mathrm{~m}$ Frequency $(\mathrm{f})=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456 \mathrm{~Hz}$ Wavelength $(\lambda)=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{2}=3.14 \mathrm{~m} \square 3.518 \mathrm{~m}$
