Question

On the superposition of two harmonic oscillations represented by${x_1} = a\sin \left( {\omega t + {\phi _1}} \right)$ and ${x_2} = a\sin \left( {\omega t + {\phi _2}} \right)$ a resulting oscillation with the same time period and amplitude is obtained. The value of ${\phi _1} - {\phi _2}$ is:
A. $120^\circ$
B. $90^\circ$
C. $60^\circ$
D. $15^\circ$

Answer 1

Recall the formula for the resulting amplitude of the wave resulted from the superposition of the two waves. Since the amplitude of resultant wave and the two super positioned waves is the same, calculate the difference in the phase angles. The cosine function is negative in the second and third quadrant.

Formula used:
$A = \sqrt {a_1^2 + a_2^2 + 2{a_1}{a_2}\cos \left( {{\phi _1} - {\phi _2}} \right)}$
where, ${a_1}$ is the amplitude of the first wave, ${a_2}$ is the amplitude of the second wave, ${\phi _1}$ and ${\phi _2}$ are the phase angles of the two waves respectively.

Complete step by step answer:
We have given the displacement equations of two harmonic oscillations. The amplitude of these two harmonic oscillations is the same. We have the expression for the resulting amplitude of the wave resulted from the superposition of the two waves,
$A = \sqrt {a_1^2 + a_2^2 + 2{a_1}{a_2}\cos \left( {{\phi _1} - {\phi _2}} \right)}$

Here, ${a_1}$ is the amplitude of the first wave, ${a_2}$ is the amplitude of the second wave, ${\phi _1}$ and ${\phi _2}$ are the phase angles of the two waves respectively.

Since we have given that the amplitude of the resultant waves is the same as that of the two waves, we can substitute ${a_1} = {a_2} = A = a$ in the above equation.
$a = \sqrt {{a^2} + {a^2} + 2{a^2}\cos \left( {{\phi _1} - {\phi _2}} \right)}$
$\Rightarrow a = \sqrt {2{a^2} + 2{a^2}\cos \left( {{\phi _1} - {\phi _2}} \right)}$
Squaring the above equation, we get,
${a^2} = 2{a^2} + 2{a^2}\cos \left( {{\phi _1} - {\phi _2}} \right)$
$\Rightarrow - {a^2} = 2{a^2}\cos \left( {{\phi _1} - {\phi _2}} \right)$
$\Rightarrow \cos \left( {{\phi _1} - {\phi _2}} \right) = - \dfrac{1}{2}$
$\therefore {\phi _1} - {\phi _2} = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$
As we know, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = 120^\circ$.
Therefore, the value of ${\phi _1} - {\phi _2}$ is $120^\circ$.

So, the correct answer is option A.

Note: To calculate the value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$, we have used the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = 60^\circ$. Since the cos function is negative in the second and third quadrant, for the second quadrant, $\theta = 180^\circ - 60^\circ = 120^\circ$. The cos is also negative in the third quadrant. Therefore, $\theta = 180^\circ + 60^\circ = 240^\circ$ in the third quadrant.