MHT CET · Physics · Oscillations
The bob of simple pendulum of length ' \(L\) ' is released from a position of small angular displacement \(\theta\). Its linear displacement at time ' \(\mathrm{t}\) ' is ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]\)
- B \(\mathrm{L} \theta \sin \left[2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \cdot \mathrm{t}\right]\)
- C \(\mathrm{L} \theta \cos \left[2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \cdot \mathrm{t}\right]\)
- D \(L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]\)
Answer & Solution
Correct Answer
(A) \(L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]\)
Step-by-step Solution
Detailed explanation
Equation for displacement for a particle performing S.H.M. is,
\(
\mathrm{y}=\mathrm{A} \cos \omega \mathrm{t}=\mathrm{L} \cos \left(\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}\right)
\)
But time period of a simple pendulum,
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}
\)
\(\therefore \quad y=L \cos \left(\frac{2 \pi}{2 \pi \sqrt{\frac{L}{g}} \times t}\right)=L \cos \left(\sqrt{\frac{g}{L}} \times t\right)\)
\(\therefore \quad\) The linear displacement is,
\(
\mathrm{s}=\mathrm{y} \theta=L \theta \cos \left[\sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}\right]
\)
\(
\mathrm{y}=\mathrm{A} \cos \omega \mathrm{t}=\mathrm{L} \cos \left(\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}\right)
\)
But time period of a simple pendulum,
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}
\)
\(\therefore \quad y=L \cos \left(\frac{2 \pi}{2 \pi \sqrt{\frac{L}{g}} \times t}\right)=L \cos \left(\sqrt{\frac{g}{L}} \times t\right)\)
\(\therefore \quad\) The linear displacement is,
\(
\mathrm{s}=\mathrm{y} \theta=L \theta \cos \left[\sqrt{\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}\right]
\)
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