Single Choice

For the travelling harmonic wave $$y(x,t)=2.0 cos $$ $$ 2\pi $$ (10t-0.0080 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 $$x$$

A$$x=4 m,\ \ \Delta\phi=6.4π \ rad $$
Correct Answer
B$$0.5 m,\ \ \ \ \ \Delta\phi=0.6π \, rad $$
C$$ \displaystyle \lambda /2 ,\ \ \ \ \ \ \ \Delta\phi= .6π \ rad$$
D$$ \displaystyle 3\lambda /4,\ \ \ \ \ \Delta\phi= 2.5π \ rad .$$

Solution

Equation for a travelling harmonic wave is given as:
$$y(x, t)=2.0\,cos\,2\pi(10t-0.0080x+0.35)$$
$$=2.0\,cos(20\pi t-0.016\pi x+0.70\pi)$$
Where,
Propagation constant, $$k = 0.0160\pi$$
Amplitude, $$a=2\,cm$$
Angular frequency, $$\omega =20\pi\,rad/s$$
Phase difference is given by the relation:
$$\phi =kx=2\pi/\lambda$$

(a) For $$\Delta x=4m= 400 cm$$
$$\Delta \phi = 0.016\pi\times 400=6.4\pi\, rad$$

(b) For $$\Delta x=0.5 m = 50 cm$$
$$\Delta \phi = 0.016\pi \times 50 = 0.8\pi\, rad$$

(c) For $$\Delta x=\lambda/2$$
$$\Delta \phi=2\pi/\lambda \times \lambda/2=\pi\, rad$$

(d) For $$\Delta x=3\lambda/4$$
$$\Delta \phi=2\pi/\lambda \times 3\lambda/4=1.5\pi\, rad$$.

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