Question 420: Waves

Question

Two waves are represented by the equation, $$\mathrm{y}_1=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx}+0.57) \mathrm{m}$$ and $$\mathrm{y}_2=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}) \mathrm{m}$$, where $$\mathrm{x}$$ is in metre and $$\mathrm{t}$$ is in second. What is the phase difference between them?

Accepted Answer

To determine the phase difference between the two waves represented by the equations:
$$\mathrm{y}_1 = \mathrm{A} \sin (\omega \mathrm{t} + \mathrm{kx} + 0.57) \mathrm{m} $$
and
$$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) \mathrm{m} $$
we need to express both waves in a similar trigonometric form to compare their phases directly. The cosine function can be rewritten in terms of the sine function using the identity:
$$\cos(	heta) = \sin\left(	heta + \frac{\pi}{2}ight)$$
Hence, we can rewrite $$\mathrm{y}_2$$ as follows:
$$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} ight) $$
Now, compare $$\mathrm{y}_1$$ and the rewritten form of $$\mathrm{y}_2$$:
$$\mathrm{y}_1 = \mathrm{A} \sin (\omega \mathrm{t} + \mathrm{kx} + 0.57) \mathrm{m}$$
$$\mathrm{y}_2 = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} ight) \mathrm{m}$$
The wave $$\mathrm{y}_1$$ has a phase term of $$(\omega \mathrm{t} + \mathrm{kx} + 0.57)$$, and the rewritten wave $$\mathrm{y}_2$$ has a phase term of $$(\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2})$$.
Thus, the phase difference between the two waves, $$\phi$$, is the difference between these two phase terms:
$$\phi = \left( (\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2}) ight) - \left( \omega \mathrm{t} + \mathrm{kx} + 0.57 ight) $$
$$\phi = \frac{\pi}{2} - 0.57 $$
Now, we need to evaluate $$\frac{\pi}{2}$$ in radians. Recall that $$\pi \approx 3.14159265$$. Therefore,
$$\frac{\pi}{2} \approx \frac{3.14159265}{2} \approx 1.57 \, 	ext{radians}$$
Substituting this value, we get:
$$\phi = 1.57 - 0.57 = 1.0 \, 	ext{radians}$$
Therefore, the phase difference between the two waves is:
Option B
1.0 radian

Answer

0.57 radian

Answer

1.0 radian

Answer

1.57 radian

Answer

1.25 radian

Waves

Two waves are represented by the equation, $\mathrm{y}_1=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx}+0.57) \mathrm{m} and \mathrm{y}_2=\mathrm{A} \cos (\omega \mathrm{t}+\mathrm{kx}) \mathrm{m}, where \mathrm{x} is in metre and \mathrm{t}$ is in second. What is the phase difference between them?

Choose the correct answer: