MHT CET · Physics · Oscillations
The displacements of two particles executing simple harmonic motion are represented as \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and \(\mathrm{y}_{2}=3 \cos 10 \mathrm{t} .\) The phase difference
between the velocities of these waves is
- A \(\left(\theta+\frac{\pi}{2}\right)\)
- B \(-\theta\)
- C \(\left(\theta-\frac{\pi}{2}\right)\)
- D \(\theta\)
Answer & Solution
Correct Answer
(C) \(\left(\theta-\frac{\pi}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{y}_{1} =2 \sin (10 \mathrm{t}+\theta) \quad \therefore \mathrm{V}_{1}=\frac{\mathrm{dy}_{1}}{\mathrm{dt}}=20 \cos\)\( (10 \mathrm{t}+\theta) \)
\( \mathrm{y}_{2} =3 \cos 10 \mathrm{t} \)
\( =-30 \cos \left(\frac{\pi}{2}-10 \mathrm{t}\right) \)
\( =-30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}\right) \)
\( =30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}+\pi\right) \)
\( =30 \cos \left(10 \mathrm{t}+\frac{\pi}{2}\right)\)
\(\therefore\) phase difference between \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) \(=(10 t+\theta)-\left(10 t+\frac{\pi}{2}\right)=\left(\theta-\frac{\pi}{2}\right)\)
\( \mathrm{y}_{2} =3 \cos 10 \mathrm{t} \)
\( =-30 \cos \left(\frac{\pi}{2}-10 \mathrm{t}\right) \)
\( =-30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}\right) \)
\( =30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}+\pi\right) \)
\( =30 \cos \left(10 \mathrm{t}+\frac{\pi}{2}\right)\)
\(\therefore\) phase difference between \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) \(=(10 t+\theta)-\left(10 t+\frac{\pi}{2}\right)=\left(\theta-\frac{\pi}{2}\right)\)
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