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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations1 Mark
MCQ
The displacement of a particle varies with time according to the relation : $\text{y}=\text{a}\sin\omega\text{t}+\text{b}\cos\omega\text{t}$
  • A
    The motion is oscillatory but not $\text{S.H.M.}$
  • B
    The motion is $\text{S.H.M.}$ with amplitude $a + b.$
  • C
    The motion is $\text{S.H.M.}$ with amplitude $\text{a}^2+\text{b}^2$
  • ✓
    The motion is $\text{S.H.M.}$ with amplitude $\sqrt{\text{a}^2+\text{b}^2}.$
✓

Answer

Correct option: D.
The motion is $\text{S.H.M.}$ with amplitude $\sqrt{\text{a}^2+\text{b}^2}.$
key concept: The sum of two $\text{S.H.Ms}$ of same frequencies is a $\text{S.H.M.}$
According to the question, the displacement
$\text{y}=\text{a}\sin\omega\text{t}+\text{b}\cos\omega\text{t}$
Given $\text{x}=\text{a}\sin\omega\text{t}+\text{b}\cos\omega\text{t}$
Let $\text{a}=\text{A}\cos\phi$
And $\text{b}=\text{A}\sin\phi$
Squaring and adding $(ii)$ and $(iii)$, we get
$\text{a}^2+\text{b}^2=\text{A}^2\cos^2\phi+\text{A}^2\sin^2\phi=\text{A}^2$
$=\text{A}^2\Rightarrow\text{A}=\sqrt{\text{a}^2+\text{b}^2}$
$\text{y}=\text{A}\sin\phi.\sin\omega\text{t}+\text{A}\cos\phi.\cos\omega\text{t}$
$=\text{A}\sin(\omega\text{t}+\phi)$
$\frac{\text{dy}}{\text{dt}}=\text{A}\omega\cos(\omega\text{t}+\phi)$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{A}\omega^2\sin(\omega\text{t}+\phi)$
$=-\text{A}\text{y}\omega^2=(-\text{A}\omega^2)\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}\propto(-\text{y})$
Hence, it is an equation of $\text{SHM}$ with amplitude
$\text{A}=\sqrt{\text{a}^2+\text{b}^2}$

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