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The wave described by $y = 0.25\sin(10\pi x - 2\pi t)$, where $x$ and $y$ are in metres and $t$ in seconds, is a wave travelling along the -
A
+ve x direction with frequency 1 Hz and wavelength $\lambda = 0.2$ m
B
-ve x direction with amplitude 0.25 m and wavelength $\lambda = 0.2$ m
C
-ve x direction with frequency 1 Hz
D
+ve x direction with frequency $\pi$ Hz and wavelength $\lambda = 0.2$ m
Detailed Solution
Given $y = 0.25\sin(10\pi x - 2\pi t)$
Compare with the standard equation of a progressive wave $y = A\sin(kx - \omega t)$:
$A = 0.25$ m, $k = 10\pi$ rad/m, $\omega = 2\pi$ rad/s
$k = \dfrac{2\pi}{\lambda} = 10\pi \Rightarrow \lambda = \dfrac{2\pi}{10\pi} = \dfrac{1}{5} = 0.2$ m
$\omega = 2\pi f = 2\pi \Rightarrow f = 1$ Hz
The coefficients of $t$ and $x$ have opposite signs, which means $\dfrac{dx}{dt} = v \gt 0$, i.e. the wave is propagating in the direction of increasing $x$.
So it is a wave travelling along the +ve x direction with frequency 1 Hz and wavelength 0.2 m.
Compare with the standard equation of a progressive wave $y = A\sin(kx - \omega t)$:
$A = 0.25$ m, $k = 10\pi$ rad/m, $\omega = 2\pi$ rad/s
$k = \dfrac{2\pi}{\lambda} = 10\pi \Rightarrow \lambda = \dfrac{2\pi}{10\pi} = \dfrac{1}{5} = 0.2$ m
$\omega = 2\pi f = 2\pi \Rightarrow f = 1$ Hz
The coefficients of $t$ and $x$ have opposite signs, which means $\dfrac{dx}{dt} = v \gt 0$, i.e. the wave is propagating in the direction of increasing $x$.
So it is a wave travelling along the +ve x direction with frequency 1 Hz and wavelength 0.2 m.