three shos of equal amplitude a and equal time period combine in the same direction. the difference in phase between successive shos is 60° ahead of the other. the amplitude of the resultant oscillation is 1)a 2)2a 3)0 4)4a

Question

Accepted Answer

# Explanation:
## Step1: Represent the three simple harmonic oscillations (SHOs) as vectors
Let the first SHO be $y_1 = A\sin(\omega t)$, the second $y_2=A\sin(\omega t + 60^{\circ})$ and the third $y_3 = A\sin(\omega t+120^{\circ})$. Using the phasor (vector) method for addition of simple - harmonic motions. The resultant amplitude $R$ of vectors $A_1$, $A_2$, $A_3$ is given by $R=\sqrt{A_1^{2}+A_2^{2}+A_3^{2}+2A_1A_2\cos\phi_{12}+2A_2A_3\cos\phi_{23}+2A_3A_1\cos\phi_{31}}$ where $\phi_{ij}$ is the phase difference between $i$ - th and $j$ - th phasor. Here $A_1 = A_2=A_3 = A$, $\phi_{12}=60^{\circ}$, $\phi_{23}=60^{\circ}$, $\phi_{31}=120^{\circ}$

## Step2: Substitute the values into the formula
$$
\begin{align*}
R&=\sqrt{A^{2}+A^{2}+A^{2}+2A\times A\times\cos60^{\circ}+2A\times A\times\cos60^{\circ}+2A\times A\times\cos120^{\circ}}\
&=\sqrt{3A^{2}+2A^{2}\times\frac{1}{2}+2A^{2}\times\frac{1}{2}+2A^{2}\times(-\frac{1}{2})}\
&=\sqrt{3A^{2}+A^{2}+A^{2}-A^{2}}\
&=\sqrt{4A^{2}}
\end{align*}
$$

# Answer:
$2A$ (Option 2)

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QUESTION IMAGE

Question

Explanation:

Step1: Represent the three simple harmonic oscillations (SHOs) as vectors

Step2: Substitute the values into the formula

Answer: