MHT CET · Physics · Oscillations
A particle starts oscillating simple harmonically from its equilibrium position with time period ' \(\mathrm{T}\) '. At time \(\mathrm{t}=\frac{\mathrm{T}}{12}\), the ratio of its kinetic energy to potential energy is
\(\left[\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}\right]\)
- A \(1: 4\)
- B \(3: 1\)
- C \(2: 1\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(B) \(3: 1\)
Step-by-step Solution
Detailed explanation
Take equation of motion as \(\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}\),
we have \(\mathrm{t}=\frac{\mathrm{T}}{12}\)
\(x\left(\frac{T}{12}\right)=\sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)=\operatorname{asin}\left(\frac{\pi}{6}\right)=\frac{a}{2}\)
Then the ratio of the kinetic energy to the potential energy at \(x=\frac{a}{2}\) will be given by
\(\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} k\left(a^2-x^2\right)}{\frac{1}{2} k x^2}=\frac{a^2-\left(\frac{a}{2}\right)^2}{\left(\frac{a}{2}\right)^2}=\frac{3}{1}\)
we have \(\mathrm{t}=\frac{\mathrm{T}}{12}\)
\(x\left(\frac{T}{12}\right)=\sin \left(\frac{2 \pi}{T} \times \frac{T}{12}\right)=\operatorname{asin}\left(\frac{\pi}{6}\right)=\frac{a}{2}\)
Then the ratio of the kinetic energy to the potential energy at \(x=\frac{a}{2}\) will be given by
\(\frac{\text { K.E. }}{\text { P.E. }}=\frac{\frac{1}{2} k\left(a^2-x^2\right)}{\frac{1}{2} k x^2}=\frac{a^2-\left(\frac{a}{2}\right)^2}{\left(\frac{a}{2}\right)^2}=\frac{3}{1}\)
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