Question

A wave equation is given as $y=\cos (500t-70x)$, where $y$ is in $mm$, $x$ in $m$ and $t$ is in $sec.$

• A. The speed of the wave is $\frac{7}{50}m/s$
• B. The speed of the wave is $\frac{50}{7}m/s$
• C. The frequency of oscillations is $1000\pi Hz$
• D. Two closest points which are in the same phase have separation $\frac{20\pi }{7}cm.$

Answer 1

Answer: (B), (D)

The correct options are
B The speed of the wave is $\frac{50}{7}m/s$
D Two closest points which are in the same phase have separation $\frac{20\pi }{7}cm.$

We have the wave equation as $y=\cos (500t-70x)=\sin (500t-70x+\frac{\pi }{2}))$.
This equation is of the form $y=A\sin (\omega t-kx)$ which is the equation for the transverse propogating wave.
This equation can be written as
$y=A\sin k(vt-x)$ (here v is the velocity of the wave)
Taking $k$ out we get $v=50/7$ m/s
Also the given wave can be written as
$y=A\sin 2\pi (\frac{t}{T}-\frac{x}{\lambda })$
Thus we get the distance between the two closest points which are in same phase i.e. the wavelength as $\frac{2\pi }{70}m=\frac{20\pi }{7}cm$