MCQ
Consider two SHMs along the same straight line $x_1=A_1 \sin \left(\omega t+\phi_1\right), x_2=A_2 \sin \left(\omega t+\phi_2\right)$, where $A_1$ and $A_2$ are their amplitudes and $\phi_1$ and $\phi_2$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II.
Column-IColumn-II
A.The two SHMs are in phase, $A_1=A_2=A$I.$R=A_1+A_2$
B.The two SHMs are in phase, $A_1 \neq A_2$II.$R=0$
C.The two SHMs are $90^{\circ}$ out of phase, $A_1=A_2=A$III.$R=2 A$
D.The two SHMs are $180^{\circ}$out of phase, $A_1=A_2$IV.$R=\sqrt{2} A$
  • A-III, B-I, C-IV, D-II
  • B
    A-IV, B-III, C-II, D-I
  • C
    A-I, B-III, C-II, D-IV
  • D
    A-III, B-IV, C-I, D-II

Answer

Correct option: A.
A-III, B-I, C-IV, D-II
(a) : $x_1=A_1 \sin \left(\omega t+\phi_1\right)$
$
x_2=A_2 \sin \left(\omega t+\phi_2\right)
$
where $A_1, A_2$ are amplitudes of two SHM's and $\phi_1, \phi_2$ are phase angle.
Resultant amplitude,
$
R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \left(\phi_1-\phi_2\right)}...(i)
$
(i) When two SHM's are in phase, $A_1=A_2=A$, then from equation (i)
$
R=\sqrt{A^2+A^2+2 A^2}=2 A
$
(ii) When two SHM's are in phase but $A_1 \neq A_2$, then
$
\begin{aligned}
& R=\sqrt{A_1^2+A_2^2+2 A_1 A_2}=\sqrt{\left(A_1+A_2\right)^2} \\
& R=A_1+A_2
\end{aligned}
$
(iii) When to SHMs are $90^{\circ}$ out of phase, then
$
R=\sqrt{A^2+A^2+0}=\sqrt{2} A
$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_1=A_2$,
$
\begin{aligned}
& R=\sqrt{A_1^2+A_1^2-2 A_1^2} \\
& R=0
\end{aligned}
$
Hence, option (a) is correct answer.

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