354511
The position of a transverse wave travelling in medium along positive \(x\)-axis is shown in figure at time \(t=0\). Speed of wave is \(v = 200\;m/s\). Equation of the wave is (in SI unit)
1 \(y=0.04 \sin 2 \pi\left(5 x-10^{3} t\right)\)
2 \(y=0.04 \sin 2 \pi\left(10^{3} t-5 x\right)\)
3 \(y=0.04 \cos 2 \pi\left(5 x-10^{3} t\right)\)
4 \(y=0.04 \cos 2 \pi\left(10^{3} t-5 x\right)\)
Explanation:
\(a=\) amplitude \(=0.04 m, \lambda=0.2 m\) \(\therefore \omega = \frac{{2\pi v}}{\lambda } = \frac{{2\pi \times 200}}{{0.2}}rad/s = 2\pi \times {10^3}rad/s\). The correct option is (1)
PHXI15:WAVES
354512
The equation of the progressive wave is \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\). The ratio of maximum particle velocity to wave velocity is
1 \(\dfrac{\pi a}{5}\)
2 \(\dfrac{2 \pi a}{5}\)
3 \(\dfrac{3 \pi a}{5}\)
4 \(\dfrac{4 \pi a}{5}\)
Explanation:
We have \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\) or \(y=a \sin 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right)\left[\because n=\dfrac{v}{\lambda}\right]\) Here, \(v\) is wave velocity \(\begin{aligned}& u=\dfrac{d y}{d t}=\dfrac{2 \pi a v}{\lambda} \cos 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right) \\& u_{\text {max }}=\dfrac{2 \pi a v}{\lambda}\end{aligned}\) \(\begin{aligned}& \therefore \dfrac{u_{\max }}{v}=\dfrac{2 \pi a}{\lambda}=\dfrac{2 \pi a}{5} \\& {\left[\begin{array}{l}\because k=\dfrac{2 \pi}{\lambda} \\\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{2 \pi / 5}=5\end{array}\right]}\end{aligned}\)
MHTCET - 2019
PHXI15:WAVES
354513
Two waves are represented bt the equations \(y_{1}=a \sin (\omega t+k x+0.57) m\) and \(y_{2}=a \cos (\omega t+k x) m\) where \(x\) is in metre and \(t\) in second. The phase difference between them is
354514
A wave equation which gives the displacement along the direction is given by \(y=0.001 \sin (100 t+x)\), where \(x\) and \(y\) are in metre and \(t\) is in second. This equation represents a wave
1 travelling with a velocity of \(100\;m/s\) in the negative \(x\)-direction
2 travelling with a velocity of \(50/\pi \,m/s\) in the positive \(x\)-direction
3 of wavelength \(1\;\,m\)
4 of frequency \(\frac{{100}}{\pi }Hz\)
Explanation:
The standard equation of a wave travelling with amplitude \(a\) in the negative \(x\)-direction with angular velocity \(\omega\) is given by \(\begin{equation*}y=a \sin (\omega t+k x) \tag{1}\end{equation*}\) where, \(k\) is a wave number. We compare the given equation with the standard equation, where \(a = 0.001\;\,m,\) \(\omega = 100{\mkern 1mu} {\mkern 1mu} rad{s^{ - 1}}\) and \(k=1\) Velocity, \(v = \frac{\omega }{k} = \frac{{100}}{1} = 100\;m/s\) Wavelength, \(\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{1}=2 \pi\) Frequency, \(v=\dfrac{\omega}{2 \pi}=\dfrac{100}{2 \pi}\) Rest all statements are incorrect.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
354511
The position of a transverse wave travelling in medium along positive \(x\)-axis is shown in figure at time \(t=0\). Speed of wave is \(v = 200\;m/s\). Equation of the wave is (in SI unit)
1 \(y=0.04 \sin 2 \pi\left(5 x-10^{3} t\right)\)
2 \(y=0.04 \sin 2 \pi\left(10^{3} t-5 x\right)\)
3 \(y=0.04 \cos 2 \pi\left(5 x-10^{3} t\right)\)
4 \(y=0.04 \cos 2 \pi\left(10^{3} t-5 x\right)\)
Explanation:
\(a=\) amplitude \(=0.04 m, \lambda=0.2 m\) \(\therefore \omega = \frac{{2\pi v}}{\lambda } = \frac{{2\pi \times 200}}{{0.2}}rad/s = 2\pi \times {10^3}rad/s\). The correct option is (1)
PHXI15:WAVES
354512
The equation of the progressive wave is \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\). The ratio of maximum particle velocity to wave velocity is
1 \(\dfrac{\pi a}{5}\)
2 \(\dfrac{2 \pi a}{5}\)
3 \(\dfrac{3 \pi a}{5}\)
4 \(\dfrac{4 \pi a}{5}\)
Explanation:
We have \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\) or \(y=a \sin 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right)\left[\because n=\dfrac{v}{\lambda}\right]\) Here, \(v\) is wave velocity \(\begin{aligned}& u=\dfrac{d y}{d t}=\dfrac{2 \pi a v}{\lambda} \cos 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right) \\& u_{\text {max }}=\dfrac{2 \pi a v}{\lambda}\end{aligned}\) \(\begin{aligned}& \therefore \dfrac{u_{\max }}{v}=\dfrac{2 \pi a}{\lambda}=\dfrac{2 \pi a}{5} \\& {\left[\begin{array}{l}\because k=\dfrac{2 \pi}{\lambda} \\\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{2 \pi / 5}=5\end{array}\right]}\end{aligned}\)
MHTCET - 2019
PHXI15:WAVES
354513
Two waves are represented bt the equations \(y_{1}=a \sin (\omega t+k x+0.57) m\) and \(y_{2}=a \cos (\omega t+k x) m\) where \(x\) is in metre and \(t\) in second. The phase difference between them is
354514
A wave equation which gives the displacement along the direction is given by \(y=0.001 \sin (100 t+x)\), where \(x\) and \(y\) are in metre and \(t\) is in second. This equation represents a wave
1 travelling with a velocity of \(100\;m/s\) in the negative \(x\)-direction
2 travelling with a velocity of \(50/\pi \,m/s\) in the positive \(x\)-direction
3 of wavelength \(1\;\,m\)
4 of frequency \(\frac{{100}}{\pi }Hz\)
Explanation:
The standard equation of a wave travelling with amplitude \(a\) in the negative \(x\)-direction with angular velocity \(\omega\) is given by \(\begin{equation*}y=a \sin (\omega t+k x) \tag{1}\end{equation*}\) where, \(k\) is a wave number. We compare the given equation with the standard equation, where \(a = 0.001\;\,m,\) \(\omega = 100{\mkern 1mu} {\mkern 1mu} rad{s^{ - 1}}\) and \(k=1\) Velocity, \(v = \frac{\omega }{k} = \frac{{100}}{1} = 100\;m/s\) Wavelength, \(\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{1}=2 \pi\) Frequency, \(v=\dfrac{\omega}{2 \pi}=\dfrac{100}{2 \pi}\) Rest all statements are incorrect.
354511
The position of a transverse wave travelling in medium along positive \(x\)-axis is shown in figure at time \(t=0\). Speed of wave is \(v = 200\;m/s\). Equation of the wave is (in SI unit)
1 \(y=0.04 \sin 2 \pi\left(5 x-10^{3} t\right)\)
2 \(y=0.04 \sin 2 \pi\left(10^{3} t-5 x\right)\)
3 \(y=0.04 \cos 2 \pi\left(5 x-10^{3} t\right)\)
4 \(y=0.04 \cos 2 \pi\left(10^{3} t-5 x\right)\)
Explanation:
\(a=\) amplitude \(=0.04 m, \lambda=0.2 m\) \(\therefore \omega = \frac{{2\pi v}}{\lambda } = \frac{{2\pi \times 200}}{{0.2}}rad/s = 2\pi \times {10^3}rad/s\). The correct option is (1)
PHXI15:WAVES
354512
The equation of the progressive wave is \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\). The ratio of maximum particle velocity to wave velocity is
1 \(\dfrac{\pi a}{5}\)
2 \(\dfrac{2 \pi a}{5}\)
3 \(\dfrac{3 \pi a}{5}\)
4 \(\dfrac{4 \pi a}{5}\)
Explanation:
We have \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\) or \(y=a \sin 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right)\left[\because n=\dfrac{v}{\lambda}\right]\) Here, \(v\) is wave velocity \(\begin{aligned}& u=\dfrac{d y}{d t}=\dfrac{2 \pi a v}{\lambda} \cos 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right) \\& u_{\text {max }}=\dfrac{2 \pi a v}{\lambda}\end{aligned}\) \(\begin{aligned}& \therefore \dfrac{u_{\max }}{v}=\dfrac{2 \pi a}{\lambda}=\dfrac{2 \pi a}{5} \\& {\left[\begin{array}{l}\because k=\dfrac{2 \pi}{\lambda} \\\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{2 \pi / 5}=5\end{array}\right]}\end{aligned}\)
MHTCET - 2019
PHXI15:WAVES
354513
Two waves are represented bt the equations \(y_{1}=a \sin (\omega t+k x+0.57) m\) and \(y_{2}=a \cos (\omega t+k x) m\) where \(x\) is in metre and \(t\) in second. The phase difference between them is
354514
A wave equation which gives the displacement along the direction is given by \(y=0.001 \sin (100 t+x)\), where \(x\) and \(y\) are in metre and \(t\) is in second. This equation represents a wave
1 travelling with a velocity of \(100\;m/s\) in the negative \(x\)-direction
2 travelling with a velocity of \(50/\pi \,m/s\) in the positive \(x\)-direction
3 of wavelength \(1\;\,m\)
4 of frequency \(\frac{{100}}{\pi }Hz\)
Explanation:
The standard equation of a wave travelling with amplitude \(a\) in the negative \(x\)-direction with angular velocity \(\omega\) is given by \(\begin{equation*}y=a \sin (\omega t+k x) \tag{1}\end{equation*}\) where, \(k\) is a wave number. We compare the given equation with the standard equation, where \(a = 0.001\;\,m,\) \(\omega = 100{\mkern 1mu} {\mkern 1mu} rad{s^{ - 1}}\) and \(k=1\) Velocity, \(v = \frac{\omega }{k} = \frac{{100}}{1} = 100\;m/s\) Wavelength, \(\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{1}=2 \pi\) Frequency, \(v=\dfrac{\omega}{2 \pi}=\dfrac{100}{2 \pi}\) Rest all statements are incorrect.
354511
The position of a transverse wave travelling in medium along positive \(x\)-axis is shown in figure at time \(t=0\). Speed of wave is \(v = 200\;m/s\). Equation of the wave is (in SI unit)
1 \(y=0.04 \sin 2 \pi\left(5 x-10^{3} t\right)\)
2 \(y=0.04 \sin 2 \pi\left(10^{3} t-5 x\right)\)
3 \(y=0.04 \cos 2 \pi\left(5 x-10^{3} t\right)\)
4 \(y=0.04 \cos 2 \pi\left(10^{3} t-5 x\right)\)
Explanation:
\(a=\) amplitude \(=0.04 m, \lambda=0.2 m\) \(\therefore \omega = \frac{{2\pi v}}{\lambda } = \frac{{2\pi \times 200}}{{0.2}}rad/s = 2\pi \times {10^3}rad/s\). The correct option is (1)
PHXI15:WAVES
354512
The equation of the progressive wave is \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\). The ratio of maximum particle velocity to wave velocity is
1 \(\dfrac{\pi a}{5}\)
2 \(\dfrac{2 \pi a}{5}\)
3 \(\dfrac{3 \pi a}{5}\)
4 \(\dfrac{4 \pi a}{5}\)
Explanation:
We have \(y=a \sin 2 \pi\left(n t-\dfrac{x}{5}\right)\) or \(y=a \sin 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right)\left[\because n=\dfrac{v}{\lambda}\right]\) Here, \(v\) is wave velocity \(\begin{aligned}& u=\dfrac{d y}{d t}=\dfrac{2 \pi a v}{\lambda} \cos 2 \pi\left(\dfrac{v t}{\lambda}-\dfrac{x}{5}\right) \\& u_{\text {max }}=\dfrac{2 \pi a v}{\lambda}\end{aligned}\) \(\begin{aligned}& \therefore \dfrac{u_{\max }}{v}=\dfrac{2 \pi a}{\lambda}=\dfrac{2 \pi a}{5} \\& {\left[\begin{array}{l}\because k=\dfrac{2 \pi}{\lambda} \\\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{2 \pi / 5}=5\end{array}\right]}\end{aligned}\)
MHTCET - 2019
PHXI15:WAVES
354513
Two waves are represented bt the equations \(y_{1}=a \sin (\omega t+k x+0.57) m\) and \(y_{2}=a \cos (\omega t+k x) m\) where \(x\) is in metre and \(t\) in second. The phase difference between them is
354514
A wave equation which gives the displacement along the direction is given by \(y=0.001 \sin (100 t+x)\), where \(x\) and \(y\) are in metre and \(t\) is in second. This equation represents a wave
1 travelling with a velocity of \(100\;m/s\) in the negative \(x\)-direction
2 travelling with a velocity of \(50/\pi \,m/s\) in the positive \(x\)-direction
3 of wavelength \(1\;\,m\)
4 of frequency \(\frac{{100}}{\pi }Hz\)
Explanation:
The standard equation of a wave travelling with amplitude \(a\) in the negative \(x\)-direction with angular velocity \(\omega\) is given by \(\begin{equation*}y=a \sin (\omega t+k x) \tag{1}\end{equation*}\) where, \(k\) is a wave number. We compare the given equation with the standard equation, where \(a = 0.001\;\,m,\) \(\omega = 100{\mkern 1mu} {\mkern 1mu} rad{s^{ - 1}}\) and \(k=1\) Velocity, \(v = \frac{\omega }{k} = \frac{{100}}{1} = 100\;m/s\) Wavelength, \(\lambda=\dfrac{2 \pi}{k}=\dfrac{2 \pi}{1}=2 \pi\) Frequency, \(v=\dfrac{\omega}{2 \pi}=\dfrac{100}{2 \pi}\) Rest all statements are incorrect.
