A pendulum made of a uniform wire of cross-sectional area \(A\) has time period \(T\). When an additional mass \(M\) is added to its bob, the time period changes to \(T_M\). If the Young’s modulus of the material of the wire is \(Y\) then \(\frac{1}{Y}\) is equal to:
(\(g=\) gravitational acceleration)
1. \( \left[\left(\frac{{T}_{{M}}}{{T}}\right)^2-1\right] \frac{{Mg}}{{A}} \)
2. \(\left[1-\left(\frac{{T}_{{M}}}{{T}}\right)^2\right] \frac{{A}}{{Mg}} \)
3. \(\left[1-\left(\frac{{T}}{{T}_{{M}}}\right)^2\right] \frac{{A}}{{Mg}} \)
4. \(\left[\left(\frac{{T}_{{M}}}{{T}}\right)^2-1\right] \frac{{A}}{{Mg}}\)

For a simple pendulum, a graph is plotted between its kinetic energy (\(KE\)) and potential energy (\(PE\)) against its displacement \(d\). Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

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A particle performs simple harmonic motion with amplitude \(A\). Its speed is tripled at the instant that it is at a distance \(\frac{2A}{3}\) from the equilibrium position. The new amplitude of the motion is: 
1. \( \frac{A}{3} \sqrt{41} \)
2. \(3 \mathrm{A} \)
3. \(A \sqrt{3} \)
4. \(\frac{7 A}{3}\)

A particle executes simple harmonic motion with a time period \(T.\) It starts at its equilibrium position at \(t=0.\) How will the graph of its kinetic energy \((KE)\) versus time \((t)\) look like?

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