172171
A wave is represented by $x=0.4$ $\cos \left(8 t-\frac{y}{2}\right)$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is
1 $0.5 \mathrm{~m} \cdot \mathrm{s}^{-1}$
2 $8 \mathrm{~m} \cdot \mathrm{s}^{-1}$
3 $16 \mathrm{~m} \cdot \mathrm{s}^{-1}$
4 $0.1 \mathrm{~m} \cdot \mathrm{s}^{-1}$
Explanation:
C Equation of the wave given by $\mathrm{x}=0.4 \cos \left(8 \mathrm{t}-\frac{\mathrm{y}}{2}\right)$ Standard equation of wave $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{ky})$ On comparing $\omega=8, \mathrm{k}=\frac{1}{2}$ Speed $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{8}{1 / 2}$ $\mathrm{v}=16 \mathrm{~m} / \mathrm{sec}$ The speed wave is $16 \mathrm{~m} / \mathrm{s}$
AP EAMCET-07.10.2020
WAVES
172172
The equation of the displacement of waves is $y$ (in $\mathrm{cm})=10(\sqrt{3} \sin 2 \pi \mathrm{t}+\cos 2 \pi \mathrm{t})$ The amplitude of the wave is.
1 $10 \mathrm{~cm}$
2 $17.3 \mathrm{~cm}$
3 $20 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
C Given that, $Y=10(\sqrt{3} \sin 2 \pi t+\cos 2 \pi t)$ $Y=(10 \sqrt{3} \sin 2 \pi t+10 \cos 2 \pi t)$ The amplitude of the wave- $\mathrm{A}_{\text {resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}}$ $=\sqrt{(10 \sqrt{3})^{2}+(10)^{2}}$ $=\sqrt{300+100}$ $\mathrm{~A}_{\text {resultant }} =20 \mathrm{~cm}$
UPSEE 2020
WAVES
172173
A travelling wave in a medium is given by the equation $y=a \sin (\omega t-k x)$. The maximum acceleration of the particle in the medium is
172175
A progressive wave of frequency $50 \mathrm{~Hz}$ is travelling with velocity $350 \mathrm{~m} / \mathrm{s}$ through a medium. The change in phase at a given time interval of $0.01 \mathrm{~s}$ is
1 $\frac{3 \pi}{2} \mathrm{rad}$
2 $\frac{\pi}{4} \mathrm{rad}$
3 $\pi \mathrm{rad}$
4 $\frac{\pi}{2} \mathrm{rad}$
Explanation:
C Given that, $\text { Frequency, } \mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{t}=0.01 \mathrm{~s}$ $\because \mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{~s}$ So, phase change $\phi=\frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{2 \pi \times 0.01}{0.02}$ $\phi=\pi \mathrm{rad}$
172171
A wave is represented by $x=0.4$ $\cos \left(8 t-\frac{y}{2}\right)$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is
1 $0.5 \mathrm{~m} \cdot \mathrm{s}^{-1}$
2 $8 \mathrm{~m} \cdot \mathrm{s}^{-1}$
3 $16 \mathrm{~m} \cdot \mathrm{s}^{-1}$
4 $0.1 \mathrm{~m} \cdot \mathrm{s}^{-1}$
Explanation:
C Equation of the wave given by $\mathrm{x}=0.4 \cos \left(8 \mathrm{t}-\frac{\mathrm{y}}{2}\right)$ Standard equation of wave $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{ky})$ On comparing $\omega=8, \mathrm{k}=\frac{1}{2}$ Speed $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{8}{1 / 2}$ $\mathrm{v}=16 \mathrm{~m} / \mathrm{sec}$ The speed wave is $16 \mathrm{~m} / \mathrm{s}$
AP EAMCET-07.10.2020
WAVES
172172
The equation of the displacement of waves is $y$ (in $\mathrm{cm})=10(\sqrt{3} \sin 2 \pi \mathrm{t}+\cos 2 \pi \mathrm{t})$ The amplitude of the wave is.
1 $10 \mathrm{~cm}$
2 $17.3 \mathrm{~cm}$
3 $20 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
C Given that, $Y=10(\sqrt{3} \sin 2 \pi t+\cos 2 \pi t)$ $Y=(10 \sqrt{3} \sin 2 \pi t+10 \cos 2 \pi t)$ The amplitude of the wave- $\mathrm{A}_{\text {resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}}$ $=\sqrt{(10 \sqrt{3})^{2}+(10)^{2}}$ $=\sqrt{300+100}$ $\mathrm{~A}_{\text {resultant }} =20 \mathrm{~cm}$
UPSEE 2020
WAVES
172173
A travelling wave in a medium is given by the equation $y=a \sin (\omega t-k x)$. The maximum acceleration of the particle in the medium is
172175
A progressive wave of frequency $50 \mathrm{~Hz}$ is travelling with velocity $350 \mathrm{~m} / \mathrm{s}$ through a medium. The change in phase at a given time interval of $0.01 \mathrm{~s}$ is
1 $\frac{3 \pi}{2} \mathrm{rad}$
2 $\frac{\pi}{4} \mathrm{rad}$
3 $\pi \mathrm{rad}$
4 $\frac{\pi}{2} \mathrm{rad}$
Explanation:
C Given that, $\text { Frequency, } \mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{t}=0.01 \mathrm{~s}$ $\because \mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{~s}$ So, phase change $\phi=\frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{2 \pi \times 0.01}{0.02}$ $\phi=\pi \mathrm{rad}$
172171
A wave is represented by $x=0.4$ $\cos \left(8 t-\frac{y}{2}\right)$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is
1 $0.5 \mathrm{~m} \cdot \mathrm{s}^{-1}$
2 $8 \mathrm{~m} \cdot \mathrm{s}^{-1}$
3 $16 \mathrm{~m} \cdot \mathrm{s}^{-1}$
4 $0.1 \mathrm{~m} \cdot \mathrm{s}^{-1}$
Explanation:
C Equation of the wave given by $\mathrm{x}=0.4 \cos \left(8 \mathrm{t}-\frac{\mathrm{y}}{2}\right)$ Standard equation of wave $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{ky})$ On comparing $\omega=8, \mathrm{k}=\frac{1}{2}$ Speed $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{8}{1 / 2}$ $\mathrm{v}=16 \mathrm{~m} / \mathrm{sec}$ The speed wave is $16 \mathrm{~m} / \mathrm{s}$
AP EAMCET-07.10.2020
WAVES
172172
The equation of the displacement of waves is $y$ (in $\mathrm{cm})=10(\sqrt{3} \sin 2 \pi \mathrm{t}+\cos 2 \pi \mathrm{t})$ The amplitude of the wave is.
1 $10 \mathrm{~cm}$
2 $17.3 \mathrm{~cm}$
3 $20 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
C Given that, $Y=10(\sqrt{3} \sin 2 \pi t+\cos 2 \pi t)$ $Y=(10 \sqrt{3} \sin 2 \pi t+10 \cos 2 \pi t)$ The amplitude of the wave- $\mathrm{A}_{\text {resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}}$ $=\sqrt{(10 \sqrt{3})^{2}+(10)^{2}}$ $=\sqrt{300+100}$ $\mathrm{~A}_{\text {resultant }} =20 \mathrm{~cm}$
UPSEE 2020
WAVES
172173
A travelling wave in a medium is given by the equation $y=a \sin (\omega t-k x)$. The maximum acceleration of the particle in the medium is
172175
A progressive wave of frequency $50 \mathrm{~Hz}$ is travelling with velocity $350 \mathrm{~m} / \mathrm{s}$ through a medium. The change in phase at a given time interval of $0.01 \mathrm{~s}$ is
1 $\frac{3 \pi}{2} \mathrm{rad}$
2 $\frac{\pi}{4} \mathrm{rad}$
3 $\pi \mathrm{rad}$
4 $\frac{\pi}{2} \mathrm{rad}$
Explanation:
C Given that, $\text { Frequency, } \mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{t}=0.01 \mathrm{~s}$ $\because \mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{~s}$ So, phase change $\phi=\frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{2 \pi \times 0.01}{0.02}$ $\phi=\pi \mathrm{rad}$
172171
A wave is represented by $x=0.4$ $\cos \left(8 t-\frac{y}{2}\right)$ where $x$ and $y$ are in meters and $t$ in secs. The speed of the wave is
1 $0.5 \mathrm{~m} \cdot \mathrm{s}^{-1}$
2 $8 \mathrm{~m} \cdot \mathrm{s}^{-1}$
3 $16 \mathrm{~m} \cdot \mathrm{s}^{-1}$
4 $0.1 \mathrm{~m} \cdot \mathrm{s}^{-1}$
Explanation:
C Equation of the wave given by $\mathrm{x}=0.4 \cos \left(8 \mathrm{t}-\frac{\mathrm{y}}{2}\right)$ Standard equation of wave $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}-\mathrm{ky})$ On comparing $\omega=8, \mathrm{k}=\frac{1}{2}$ Speed $(\mathrm{v})=\frac{\omega}{\mathrm{k}}=\frac{8}{1 / 2}$ $\mathrm{v}=16 \mathrm{~m} / \mathrm{sec}$ The speed wave is $16 \mathrm{~m} / \mathrm{s}$
AP EAMCET-07.10.2020
WAVES
172172
The equation of the displacement of waves is $y$ (in $\mathrm{cm})=10(\sqrt{3} \sin 2 \pi \mathrm{t}+\cos 2 \pi \mathrm{t})$ The amplitude of the wave is.
1 $10 \mathrm{~cm}$
2 $17.3 \mathrm{~cm}$
3 $20 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
C Given that, $Y=10(\sqrt{3} \sin 2 \pi t+\cos 2 \pi t)$ $Y=(10 \sqrt{3} \sin 2 \pi t+10 \cos 2 \pi t)$ The amplitude of the wave- $\mathrm{A}_{\text {resultant }} =\sqrt{\mathrm{A}_{1}^{2}+\mathrm{A}_{2}^{2}}$ $=\sqrt{(10 \sqrt{3})^{2}+(10)^{2}}$ $=\sqrt{300+100}$ $\mathrm{~A}_{\text {resultant }} =20 \mathrm{~cm}$
UPSEE 2020
WAVES
172173
A travelling wave in a medium is given by the equation $y=a \sin (\omega t-k x)$. The maximum acceleration of the particle in the medium is
172175
A progressive wave of frequency $50 \mathrm{~Hz}$ is travelling with velocity $350 \mathrm{~m} / \mathrm{s}$ through a medium. The change in phase at a given time interval of $0.01 \mathrm{~s}$ is
1 $\frac{3 \pi}{2} \mathrm{rad}$
2 $\frac{\pi}{4} \mathrm{rad}$
3 $\pi \mathrm{rad}$
4 $\frac{\pi}{2} \mathrm{rad}$
Explanation:
C Given that, $\text { Frequency, } \mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{t}=0.01 \mathrm{~s}$ $\because \mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{~s}$ So, phase change $\phi=\frac{2 \pi \mathrm{t}}{\mathrm{T}}=\frac{2 \pi \times 0.01}{0.02}$ $\phi=\pi \mathrm{rad}$
