354524
The equation of the progressive wave is \(y=3 \sin \left[\pi\left(\dfrac{t}{3}-\dfrac{x}{5}\right)+\dfrac{\pi}{4}\right]\), where \(x\) and \(y\) are in metre and time in second. Which of the following is correct?
1 Velocity \(v = 1.5\;m/s\)
2 Amplitude \(A = 3\;cm\)
3 Frequency \(f = 0.2\;Hz\)
4 Wavelength \(\lambda = 10\;m\)
Explanation:
Compare the given equation with the standard equation of wave motion, \(Y=A \sin \left[2 \pi\left(\dfrac{t}{T}-\dfrac{x}{\lambda}\right)+\dfrac{\pi}{4}\right]\) Where, \(A\) and \(\lambda\) are amplitude and wavelength, respectively Amplitude A \( = 3\;m\) Wavelength \(\lambda = 10\;m\).
MHTCET - 2017
PHXI15:WAVES
354525
\(y=(x, t)=\dfrac{0.8}{(4 x+5 t)^{2}+5}\) represents a moving pulse, where \(x\) and \(y\) are in metre and \(t\) is in second, then incorrect statement of the following
1 Pulse is moving in \(+x\) direction
2 In \(2\,s\) it will travel a distance of \(2.5\;m\)
3 Its maximum displacement is \(0.16\;m\)
4 It is a symmetric pulse
Explanation:
\(y=\dfrac{0.8}{(4 x+5 t)^{2}+5}=\dfrac{0.8}{4^{2}\left(x+\dfrac{5}{4} t\right)^{2}+5}\) \(y = f(n \pm vt)\) \(v = \frac{5}{4}m/s\) \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} x = vt = \frac{5}{4} \times 2 = 2.5\;m/s\) wave is moving -ve \(x\)-direction \(\therefore\) Distance traversed in \(2\,\sec = \frac{5}{4} \times 2\;m = 2.5\;m\) At \(t=0, x=0, y_{\max }(x, t)=\dfrac{0.8}{5} m=0.16 m\) At \(t=0\) and \(x= \pm a(a=a+v e\) constant ) . \(y\) will have same value. So the pulse is symmetric.
PHXI15:WAVES
354526
A wave equation which gives the displacement along the \(y\)-direction is given by \(y=10^{-4} \sin (60 t+2 x)\) where \(x\) and \(y\) are in metre and \(t\) is time in second. This represents a wave
1 Travelling with a velocity of \(30\;m/s\) in the positive \(x\)-direction
354524
The equation of the progressive wave is \(y=3 \sin \left[\pi\left(\dfrac{t}{3}-\dfrac{x}{5}\right)+\dfrac{\pi}{4}\right]\), where \(x\) and \(y\) are in metre and time in second. Which of the following is correct?
1 Velocity \(v = 1.5\;m/s\)
2 Amplitude \(A = 3\;cm\)
3 Frequency \(f = 0.2\;Hz\)
4 Wavelength \(\lambda = 10\;m\)
Explanation:
Compare the given equation with the standard equation of wave motion, \(Y=A \sin \left[2 \pi\left(\dfrac{t}{T}-\dfrac{x}{\lambda}\right)+\dfrac{\pi}{4}\right]\) Where, \(A\) and \(\lambda\) are amplitude and wavelength, respectively Amplitude A \( = 3\;m\) Wavelength \(\lambda = 10\;m\).
MHTCET - 2017
PHXI15:WAVES
354525
\(y=(x, t)=\dfrac{0.8}{(4 x+5 t)^{2}+5}\) represents a moving pulse, where \(x\) and \(y\) are in metre and \(t\) is in second, then incorrect statement of the following
1 Pulse is moving in \(+x\) direction
2 In \(2\,s\) it will travel a distance of \(2.5\;m\)
3 Its maximum displacement is \(0.16\;m\)
4 It is a symmetric pulse
Explanation:
\(y=\dfrac{0.8}{(4 x+5 t)^{2}+5}=\dfrac{0.8}{4^{2}\left(x+\dfrac{5}{4} t\right)^{2}+5}\) \(y = f(n \pm vt)\) \(v = \frac{5}{4}m/s\) \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} x = vt = \frac{5}{4} \times 2 = 2.5\;m/s\) wave is moving -ve \(x\)-direction \(\therefore\) Distance traversed in \(2\,\sec = \frac{5}{4} \times 2\;m = 2.5\;m\) At \(t=0, x=0, y_{\max }(x, t)=\dfrac{0.8}{5} m=0.16 m\) At \(t=0\) and \(x= \pm a(a=a+v e\) constant ) . \(y\) will have same value. So the pulse is symmetric.
PHXI15:WAVES
354526
A wave equation which gives the displacement along the \(y\)-direction is given by \(y=10^{-4} \sin (60 t+2 x)\) where \(x\) and \(y\) are in metre and \(t\) is time in second. This represents a wave
1 Travelling with a velocity of \(30\;m/s\) in the positive \(x\)-direction
354524
The equation of the progressive wave is \(y=3 \sin \left[\pi\left(\dfrac{t}{3}-\dfrac{x}{5}\right)+\dfrac{\pi}{4}\right]\), where \(x\) and \(y\) are in metre and time in second. Which of the following is correct?
1 Velocity \(v = 1.5\;m/s\)
2 Amplitude \(A = 3\;cm\)
3 Frequency \(f = 0.2\;Hz\)
4 Wavelength \(\lambda = 10\;m\)
Explanation:
Compare the given equation with the standard equation of wave motion, \(Y=A \sin \left[2 \pi\left(\dfrac{t}{T}-\dfrac{x}{\lambda}\right)+\dfrac{\pi}{4}\right]\) Where, \(A\) and \(\lambda\) are amplitude and wavelength, respectively Amplitude A \( = 3\;m\) Wavelength \(\lambda = 10\;m\).
MHTCET - 2017
PHXI15:WAVES
354525
\(y=(x, t)=\dfrac{0.8}{(4 x+5 t)^{2}+5}\) represents a moving pulse, where \(x\) and \(y\) are in metre and \(t\) is in second, then incorrect statement of the following
1 Pulse is moving in \(+x\) direction
2 In \(2\,s\) it will travel a distance of \(2.5\;m\)
3 Its maximum displacement is \(0.16\;m\)
4 It is a symmetric pulse
Explanation:
\(y=\dfrac{0.8}{(4 x+5 t)^{2}+5}=\dfrac{0.8}{4^{2}\left(x+\dfrac{5}{4} t\right)^{2}+5}\) \(y = f(n \pm vt)\) \(v = \frac{5}{4}m/s\) \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} x = vt = \frac{5}{4} \times 2 = 2.5\;m/s\) wave is moving -ve \(x\)-direction \(\therefore\) Distance traversed in \(2\,\sec = \frac{5}{4} \times 2\;m = 2.5\;m\) At \(t=0, x=0, y_{\max }(x, t)=\dfrac{0.8}{5} m=0.16 m\) At \(t=0\) and \(x= \pm a(a=a+v e\) constant ) . \(y\) will have same value. So the pulse is symmetric.
PHXI15:WAVES
354526
A wave equation which gives the displacement along the \(y\)-direction is given by \(y=10^{-4} \sin (60 t+2 x)\) where \(x\) and \(y\) are in metre and \(t\) is time in second. This represents a wave
1 Travelling with a velocity of \(30\;m/s\) in the positive \(x\)-direction
354524
The equation of the progressive wave is \(y=3 \sin \left[\pi\left(\dfrac{t}{3}-\dfrac{x}{5}\right)+\dfrac{\pi}{4}\right]\), where \(x\) and \(y\) are in metre and time in second. Which of the following is correct?
1 Velocity \(v = 1.5\;m/s\)
2 Amplitude \(A = 3\;cm\)
3 Frequency \(f = 0.2\;Hz\)
4 Wavelength \(\lambda = 10\;m\)
Explanation:
Compare the given equation with the standard equation of wave motion, \(Y=A \sin \left[2 \pi\left(\dfrac{t}{T}-\dfrac{x}{\lambda}\right)+\dfrac{\pi}{4}\right]\) Where, \(A\) and \(\lambda\) are amplitude and wavelength, respectively Amplitude A \( = 3\;m\) Wavelength \(\lambda = 10\;m\).
MHTCET - 2017
PHXI15:WAVES
354525
\(y=(x, t)=\dfrac{0.8}{(4 x+5 t)^{2}+5}\) represents a moving pulse, where \(x\) and \(y\) are in metre and \(t\) is in second, then incorrect statement of the following
1 Pulse is moving in \(+x\) direction
2 In \(2\,s\) it will travel a distance of \(2.5\;m\)
3 Its maximum displacement is \(0.16\;m\)
4 It is a symmetric pulse
Explanation:
\(y=\dfrac{0.8}{(4 x+5 t)^{2}+5}=\dfrac{0.8}{4^{2}\left(x+\dfrac{5}{4} t\right)^{2}+5}\) \(y = f(n \pm vt)\) \(v = \frac{5}{4}m/s\) \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} x = vt = \frac{5}{4} \times 2 = 2.5\;m/s\) wave is moving -ve \(x\)-direction \(\therefore\) Distance traversed in \(2\,\sec = \frac{5}{4} \times 2\;m = 2.5\;m\) At \(t=0, x=0, y_{\max }(x, t)=\dfrac{0.8}{5} m=0.16 m\) At \(t=0\) and \(x= \pm a(a=a+v e\) constant ) . \(y\) will have same value. So the pulse is symmetric.
PHXI15:WAVES
354526
A wave equation which gives the displacement along the \(y\)-direction is given by \(y=10^{-4} \sin (60 t+2 x)\) where \(x\) and \(y\) are in metre and \(t\) is time in second. This represents a wave
1 Travelling with a velocity of \(30\;m/s\) in the positive \(x\)-direction
