Question 511 Mark
For the travelling harmonic wave
$\text{y}(\text{x, t})=2.0\cos2\pi(10\text{t}-0.0080\text{x}+0.35)$
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
0.5m,
$\text{y}(\text{x, t})=2.0\cos2\pi(10\text{t}-0.0080\text{x}+0.35)$
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
0.5m,
Answer
View full question & answer→Equation for a travelling harmonic wave is given as:
$\text{y}(\text{x, t})=2.0\cos2\pi(10\text{t}-0.0080\text{x}+0.35)$
$=2.0\cos(20\pi\text{t}-0.016\pi\text{x}+0.70\pi)$
Where,
Propagation constant, $\text{k}=0.0160\pi$
Amplitude, a = 2cm
Angular frequency, $\omega=20\pi\text{ rad/s}$
Phase difference is given by the relation:
$\phi=\text{kx}=\frac{2\pi}{\lambda}$
For x = 0.5m = 50cm
$\phi=0.016\pi\times50=0.8\pi\text{ rad}$
$\text{y}(\text{x, t})=2.0\cos2\pi(10\text{t}-0.0080\text{x}+0.35)$
$=2.0\cos(20\pi\text{t}-0.016\pi\text{x}+0.70\pi)$
Where,
Propagation constant, $\text{k}=0.0160\pi$
Amplitude, a = 2cm
Angular frequency, $\omega=20\pi\text{ rad/s}$
Phase difference is given by the relation:
$\phi=\text{kx}=\frac{2\pi}{\lambda}$
For x = 0.5m = 50cm
$\phi=0.016\pi\times50=0.8\pi\text{ rad}$

