MHT CET · Physics · Waves and Sound
The displacement equation of a simple harmonic oscillator is given by \(y=A \sin \omega t-B \cos \omega t\)
The amplitude of the oscillator will be
- A \(A-B\)
- B \(A+B\)
- C \(\sqrt{A^{2}+B^{2}}\)
- D \(\left(A^{2}+B^{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\sqrt{A^{2}+B^{2}}\)
Step-by-step Solution
Detailed explanation
Displacement equation \(y=A \sin \omega t-B \cos \omega t\)
Let \(A=a \cos \theta\) and \(B=a \sin \theta\)
So, \(A^{2}+B^{2}=a^{2}\)
\(
a=\sqrt{A^{2}+B^{2}}
\)
Then, \(y=a \cos \theta \sin \omega t-a \sin \theta \cos \omega t\)
\(
y=a \sin (\omega t-\theta)
\)
which is the equation of simple harmonic oscillator.
The amplitude of the oscillator
\(
=a=\sqrt{A^{2}+B^{2}}
\)
Let \(A=a \cos \theta\) and \(B=a \sin \theta\)
So, \(A^{2}+B^{2}=a^{2}\)
\(
a=\sqrt{A^{2}+B^{2}}
\)
Then, \(y=a \cos \theta \sin \omega t-a \sin \theta \cos \omega t\)
\(
y=a \sin (\omega t-\theta)
\)
which is the equation of simple harmonic oscillator.
The amplitude of the oscillator
\(
=a=\sqrt{A^{2}+B^{2}}
\)
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