MHT CET · Physics · Oscillations
A bob of a simple pendulum has mass 'm' and is oscillating with an amplitude 'a'.
If the length of the pendulum is 'L' then the maximum tension in the string is
\(\left[\cos 0^{\circ}=1, \mathrm{~g}=\right.\) acceleration due to gravity \(]\)
- A \(\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]\)
- B \(\mathrm{mg}\left[1-\left(\frac{\mathrm{L}}{\mathrm{a}}\right)^{2}\right]\)
- C \(\mathrm{mg}\left[1+\left(\frac{\mathrm{L}}{\mathrm{a}}\right)^{2}\right]\)
- D \(\mathrm{mg}\left[1-\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]\)
Answer & Solution
Correct Answer
(A) \(\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]\)
Step-by-step Solution
Detailed explanation
(B)
tension in the string is maximum when the bob passes through the mean position.
\(\mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mV}^{2}}{\mathrm{~L}}\)...(1)
In S.H.M. velocity at the mean position is given by \(\mathrm{V}=\mathrm{a} \omega\) For simple pendulum \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)
\(\therefore \omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{L}}\) \(\therefore V=a \sqrt{\frac{g}{L}}\) or \(V^{2}=a^{2} \frac{g}{L}\)
Putting this value of \(\mathrm{V}^{2}\) in \(\mathrm{Eq} .(1)\) we get \(\mathrm{T}_{\max }=\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]\)
tension in the string is maximum when the bob passes through the mean position.
\(\mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mV}^{2}}{\mathrm{~L}}\)...(1)
In S.H.M. velocity at the mean position is given by \(\mathrm{V}=\mathrm{a} \omega\) For simple pendulum \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)
\(\therefore \omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{L}}\) \(\therefore V=a \sqrt{\frac{g}{L}}\) or \(V^{2}=a^{2} \frac{g}{L}\)
Putting this value of \(\mathrm{V}^{2}\) in \(\mathrm{Eq} .(1)\) we get \(\mathrm{T}_{\max }=\mathrm{mg}\left[1+\left(\frac{\mathrm{a}}{\mathrm{L}}\right)^{2}\right]\)
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