354528
The equation of longitudinal wave represented as \(y = 20\cos \pi (50t - x)\,cm,\) then its wavelength is
1 \(120\;cm\)
2 \(50\;cm\)
3 \(2\;cm\)
4 \(5\;cm\)
Explanation:
The standard equation of a wave of amplitude \(a\) is given by \(\begin{equation*}y=a \cos (\omega t-k x) \tag{1}\end{equation*}\) where, \(\omega\) is angular velocity, \(k\) is the wave number and \(t\) is the time. Comparing with given equation, we get \(k=\pi\). Wavelength, \(\lambda = \frac{{2\pi }}{k} = \frac{{2\pi }}{\pi } = 2\;cm\)
PHXI15:WAVES
354529
The position of a wave (of wavelength \(\lambda\) ) \(y(x, t)=A \sin (k x-23562 t)\) is shown at \(t = 0\), find \(x\)-coordinate of point \(P\) in metres, if the wave speed is \(300\;m/\sec {\text{ }}[\therefore y,x\) are measured in meters, \(t\) is in seconds].
1 \(\dfrac{27 \lambda}{12}\)
2 \(\dfrac{29 \lambda}{12}\)
3 \(\dfrac{25 \lambda}{12}\)
4 \(\dfrac{31 \lambda}{12}\)
Explanation:
At \(t=0, y=A \sin k x=A \sin \dfrac{2 \pi}{\lambda} x\) From the figure, \(k x=\dfrac{2 \pi}{\lambda} x=4 \pi+\dfrac{\pi}{6} \Rightarrow x=\dfrac{25 \lambda}{12}\)
PHXI15:WAVES
354530
A wave equation is given by \(y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right]\), where \(x\) is in centimetre and \(t\) is in second. Which of the following is true?
1 \(\lambda = 18\;cm\)
2 \(v = 4\,m{s^{ - 1}}\)
3 \(a = 0.4\;m\)
4 \(f = 50\;Hz\)
Explanation:
The given equation will be written as \(\begin{equation*}y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right] \tag{1}\end{equation*}\) The standard wave equation can be written as \(\begin{align*}y & =a \sin (\omega t-k x+\phi) \\\Rightarrow y & =a \sin \left(\dfrac{2 \pi}{T} t-\dfrac{2 \pi}{\lambda} \cdot x+\phi\right) \tag{2}\end{align*}\) Comparing Eqs.(1) and (2),we get Amplitude, \(a = 4\;cm = 0.04\;m\) Frequency, \(f = \frac{1}{T} = \frac{1}{{10}}\;Hz = 0.1\;Hz\) Wavelength, \(\lambda = 2 \times 9 = 18\,cm\) Velocity, \(v=f \lambda\) \( = 0.1 \times 18 = 1.8\;\,cm\;{s^{ - 1}}\)
PHXI15:WAVES
354531
A transverse wave is described by the equation \(y=y_{0} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right]\). If the maximum particle velocity is four times the wave velocity, then \(\lambda\) is equal to
1 \(y_{0}\)
2 \(\dfrac{y_{0}}{2}\)
3 \(\pi y_{0}\)
4 \(\dfrac{\pi y_{0}}{2}\)
Explanation:
The given equation is \(\begin{aligned}& y=y_{o} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right] \\& \omega=2 \pi n, k=\dfrac{2 \pi}{\lambda}\end{aligned}\) The velocity of the wave is \(v=\dfrac{\omega}{k}=n \lambda\) Given that \(v_{p}=4 v\) \(y_{o}(2 \pi n)=4 n \lambda \Rightarrow \lambda=\dfrac{\pi y_{o}}{2}\)
354528
The equation of longitudinal wave represented as \(y = 20\cos \pi (50t - x)\,cm,\) then its wavelength is
1 \(120\;cm\)
2 \(50\;cm\)
3 \(2\;cm\)
4 \(5\;cm\)
Explanation:
The standard equation of a wave of amplitude \(a\) is given by \(\begin{equation*}y=a \cos (\omega t-k x) \tag{1}\end{equation*}\) where, \(\omega\) is angular velocity, \(k\) is the wave number and \(t\) is the time. Comparing with given equation, we get \(k=\pi\). Wavelength, \(\lambda = \frac{{2\pi }}{k} = \frac{{2\pi }}{\pi } = 2\;cm\)
PHXI15:WAVES
354529
The position of a wave (of wavelength \(\lambda\) ) \(y(x, t)=A \sin (k x-23562 t)\) is shown at \(t = 0\), find \(x\)-coordinate of point \(P\) in metres, if the wave speed is \(300\;m/\sec {\text{ }}[\therefore y,x\) are measured in meters, \(t\) is in seconds].
1 \(\dfrac{27 \lambda}{12}\)
2 \(\dfrac{29 \lambda}{12}\)
3 \(\dfrac{25 \lambda}{12}\)
4 \(\dfrac{31 \lambda}{12}\)
Explanation:
At \(t=0, y=A \sin k x=A \sin \dfrac{2 \pi}{\lambda} x\) From the figure, \(k x=\dfrac{2 \pi}{\lambda} x=4 \pi+\dfrac{\pi}{6} \Rightarrow x=\dfrac{25 \lambda}{12}\)
PHXI15:WAVES
354530
A wave equation is given by \(y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right]\), where \(x\) is in centimetre and \(t\) is in second. Which of the following is true?
1 \(\lambda = 18\;cm\)
2 \(v = 4\,m{s^{ - 1}}\)
3 \(a = 0.4\;m\)
4 \(f = 50\;Hz\)
Explanation:
The given equation will be written as \(\begin{equation*}y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right] \tag{1}\end{equation*}\) The standard wave equation can be written as \(\begin{align*}y & =a \sin (\omega t-k x+\phi) \\\Rightarrow y & =a \sin \left(\dfrac{2 \pi}{T} t-\dfrac{2 \pi}{\lambda} \cdot x+\phi\right) \tag{2}\end{align*}\) Comparing Eqs.(1) and (2),we get Amplitude, \(a = 4\;cm = 0.04\;m\) Frequency, \(f = \frac{1}{T} = \frac{1}{{10}}\;Hz = 0.1\;Hz\) Wavelength, \(\lambda = 2 \times 9 = 18\,cm\) Velocity, \(v=f \lambda\) \( = 0.1 \times 18 = 1.8\;\,cm\;{s^{ - 1}}\)
PHXI15:WAVES
354531
A transverse wave is described by the equation \(y=y_{0} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right]\). If the maximum particle velocity is four times the wave velocity, then \(\lambda\) is equal to
1 \(y_{0}\)
2 \(\dfrac{y_{0}}{2}\)
3 \(\pi y_{0}\)
4 \(\dfrac{\pi y_{0}}{2}\)
Explanation:
The given equation is \(\begin{aligned}& y=y_{o} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right] \\& \omega=2 \pi n, k=\dfrac{2 \pi}{\lambda}\end{aligned}\) The velocity of the wave is \(v=\dfrac{\omega}{k}=n \lambda\) Given that \(v_{p}=4 v\) \(y_{o}(2 \pi n)=4 n \lambda \Rightarrow \lambda=\dfrac{\pi y_{o}}{2}\)
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PHXI15:WAVES
354528
The equation of longitudinal wave represented as \(y = 20\cos \pi (50t - x)\,cm,\) then its wavelength is
1 \(120\;cm\)
2 \(50\;cm\)
3 \(2\;cm\)
4 \(5\;cm\)
Explanation:
The standard equation of a wave of amplitude \(a\) is given by \(\begin{equation*}y=a \cos (\omega t-k x) \tag{1}\end{equation*}\) where, \(\omega\) is angular velocity, \(k\) is the wave number and \(t\) is the time. Comparing with given equation, we get \(k=\pi\). Wavelength, \(\lambda = \frac{{2\pi }}{k} = \frac{{2\pi }}{\pi } = 2\;cm\)
PHXI15:WAVES
354529
The position of a wave (of wavelength \(\lambda\) ) \(y(x, t)=A \sin (k x-23562 t)\) is shown at \(t = 0\), find \(x\)-coordinate of point \(P\) in metres, if the wave speed is \(300\;m/\sec {\text{ }}[\therefore y,x\) are measured in meters, \(t\) is in seconds].
1 \(\dfrac{27 \lambda}{12}\)
2 \(\dfrac{29 \lambda}{12}\)
3 \(\dfrac{25 \lambda}{12}\)
4 \(\dfrac{31 \lambda}{12}\)
Explanation:
At \(t=0, y=A \sin k x=A \sin \dfrac{2 \pi}{\lambda} x\) From the figure, \(k x=\dfrac{2 \pi}{\lambda} x=4 \pi+\dfrac{\pi}{6} \Rightarrow x=\dfrac{25 \lambda}{12}\)
PHXI15:WAVES
354530
A wave equation is given by \(y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right]\), where \(x\) is in centimetre and \(t\) is in second. Which of the following is true?
1 \(\lambda = 18\;cm\)
2 \(v = 4\,m{s^{ - 1}}\)
3 \(a = 0.4\;m\)
4 \(f = 50\;Hz\)
Explanation:
The given equation will be written as \(\begin{equation*}y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right] \tag{1}\end{equation*}\) The standard wave equation can be written as \(\begin{align*}y & =a \sin (\omega t-k x+\phi) \\\Rightarrow y & =a \sin \left(\dfrac{2 \pi}{T} t-\dfrac{2 \pi}{\lambda} \cdot x+\phi\right) \tag{2}\end{align*}\) Comparing Eqs.(1) and (2),we get Amplitude, \(a = 4\;cm = 0.04\;m\) Frequency, \(f = \frac{1}{T} = \frac{1}{{10}}\;Hz = 0.1\;Hz\) Wavelength, \(\lambda = 2 \times 9 = 18\,cm\) Velocity, \(v=f \lambda\) \( = 0.1 \times 18 = 1.8\;\,cm\;{s^{ - 1}}\)
PHXI15:WAVES
354531
A transverse wave is described by the equation \(y=y_{0} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right]\). If the maximum particle velocity is four times the wave velocity, then \(\lambda\) is equal to
1 \(y_{0}\)
2 \(\dfrac{y_{0}}{2}\)
3 \(\pi y_{0}\)
4 \(\dfrac{\pi y_{0}}{2}\)
Explanation:
The given equation is \(\begin{aligned}& y=y_{o} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right] \\& \omega=2 \pi n, k=\dfrac{2 \pi}{\lambda}\end{aligned}\) The velocity of the wave is \(v=\dfrac{\omega}{k}=n \lambda\) Given that \(v_{p}=4 v\) \(y_{o}(2 \pi n)=4 n \lambda \Rightarrow \lambda=\dfrac{\pi y_{o}}{2}\)
354528
The equation of longitudinal wave represented as \(y = 20\cos \pi (50t - x)\,cm,\) then its wavelength is
1 \(120\;cm\)
2 \(50\;cm\)
3 \(2\;cm\)
4 \(5\;cm\)
Explanation:
The standard equation of a wave of amplitude \(a\) is given by \(\begin{equation*}y=a \cos (\omega t-k x) \tag{1}\end{equation*}\) where, \(\omega\) is angular velocity, \(k\) is the wave number and \(t\) is the time. Comparing with given equation, we get \(k=\pi\). Wavelength, \(\lambda = \frac{{2\pi }}{k} = \frac{{2\pi }}{\pi } = 2\;cm\)
PHXI15:WAVES
354529
The position of a wave (of wavelength \(\lambda\) ) \(y(x, t)=A \sin (k x-23562 t)\) is shown at \(t = 0\), find \(x\)-coordinate of point \(P\) in metres, if the wave speed is \(300\;m/\sec {\text{ }}[\therefore y,x\) are measured in meters, \(t\) is in seconds].
1 \(\dfrac{27 \lambda}{12}\)
2 \(\dfrac{29 \lambda}{12}\)
3 \(\dfrac{25 \lambda}{12}\)
4 \(\dfrac{31 \lambda}{12}\)
Explanation:
At \(t=0, y=A \sin k x=A \sin \dfrac{2 \pi}{\lambda} x\) From the figure, \(k x=\dfrac{2 \pi}{\lambda} x=4 \pi+\dfrac{\pi}{6} \Rightarrow x=\dfrac{25 \lambda}{12}\)
PHXI15:WAVES
354530
A wave equation is given by \(y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right]\), where \(x\) is in centimetre and \(t\) is in second. Which of the following is true?
1 \(\lambda = 18\;cm\)
2 \(v = 4\,m{s^{ - 1}}\)
3 \(a = 0.4\;m\)
4 \(f = 50\;Hz\)
Explanation:
The given equation will be written as \(\begin{equation*}y=4 \sin \left[\pi\left(\dfrac{t}{5}-\dfrac{x}{9}+\dfrac{1}{6}\right)\right] \tag{1}\end{equation*}\) The standard wave equation can be written as \(\begin{align*}y & =a \sin (\omega t-k x+\phi) \\\Rightarrow y & =a \sin \left(\dfrac{2 \pi}{T} t-\dfrac{2 \pi}{\lambda} \cdot x+\phi\right) \tag{2}\end{align*}\) Comparing Eqs.(1) and (2),we get Amplitude, \(a = 4\;cm = 0.04\;m\) Frequency, \(f = \frac{1}{T} = \frac{1}{{10}}\;Hz = 0.1\;Hz\) Wavelength, \(\lambda = 2 \times 9 = 18\,cm\) Velocity, \(v=f \lambda\) \( = 0.1 \times 18 = 1.8\;\,cm\;{s^{ - 1}}\)
PHXI15:WAVES
354531
A transverse wave is described by the equation \(y=y_{0} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right]\). If the maximum particle velocity is four times the wave velocity, then \(\lambda\) is equal to
1 \(y_{0}\)
2 \(\dfrac{y_{0}}{2}\)
3 \(\pi y_{0}\)
4 \(\dfrac{\pi y_{0}}{2}\)
Explanation:
The given equation is \(\begin{aligned}& y=y_{o} \sin 2 \pi\left[n t-\dfrac{x}{\lambda}\right] \\& \omega=2 \pi n, k=\dfrac{2 \pi}{\lambda}\end{aligned}\) The velocity of the wave is \(v=\dfrac{\omega}{k}=n \lambda\) Given that \(v_{p}=4 v\) \(y_{o}(2 \pi n)=4 n \lambda \Rightarrow \lambda=\dfrac{\pi y_{o}}{2}\)
