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Q. No.

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. \(\left[\left(\frac{{T}_{{M}}}{{T}}\right)^2-1\right] \frac{{A}}{{Mg}}\)

Subtopic:  Angular SHM |
Level 3: 35%-60%
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A simple pendulum has a time period \(T\) in air. The bob is then completely immersed and continues to oscillate freely in a non-viscous liquid whose density is \(\left ( \dfrac{1}{16}\right )^\mathrm{th} \) of that of the bob. Assuming no damping and only the effect of buoyancy, what is the new time period of oscillation?

1. \( 2 T \sqrt{\dfrac{1}{14}} \) 2. \( 2 T \sqrt{\dfrac{1}{10}} \)
3. \(4 T \sqrt{\dfrac{1}{15}} \) 4. \( 4 T \sqrt{\dfrac{1}{14}} \)
Subtopic:  Angular SHM |
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Level 1: 80%+
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Two masses \({m}~\text{and}~ \frac {{m}}{2}\) are connected at the two ends of a massless rigid rod of length \(l.\) The rod is suspended by a thin wire of torsional constant \(k\) at the center of mass of the rod-mass system (see figure). Because of the torsional constant \(k,\) the restoring torque is \(\tau=k \theta\) for angular displacement \(\theta.\) If the rod is rotated by \(\theta_0\) and released, the tension in it when it passes through its mean position will be:
   
1. \(\dfrac{3k\theta_0^2}{l}\)

2. \(\dfrac{2k\theta_0^2}{l}\)

3. \(\dfrac{k\theta_0^2}{l}\)

4. \(\dfrac{k\theta_0^2}{2l}\)
Subtopic:  Angular SHM |
Level 3: 35%-60%
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A ring is hung on a nail and can oscillate without slipping or sliding. It can swing in its own plane with a time period \(T_1\)​ and back and forth perpendicular to its plane with a time period \(T_2.\) What is the ratio \(\dfrac{T_1}{T_2} \text{?}\)
1. \(\dfrac{2}{\sqrt{3}} \) 2. \(\dfrac{\sqrt{2}}{3} \)
3. \( \dfrac{2}{3} \) 4. \(\dfrac{3}{\sqrt{2}}\)
Subtopic:  Angular SHM |
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Level 2: 60%+
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If the time period of a \(2\) m long simple pendulum is \(2\) s, the acceleration due to gravity at the place where the pendulum is executing simple harmonic motion is:
1. \(\pi^{2}\) m/s2
2. \(2\pi^{2}\) m/s2
3. \(9.8\) m/s2
4. \(16\) m/s2

Subtopic:  Angular SHM |
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Level 2: 60%+
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Given below are two statements: 

Statement I: A seconds pendulum has a time period of \(1\) second.
Statement II: A seconds pendulum takes exactly \(1\) second to travel between its two extreme positions.
 
1. Both Statement I and Statement II are incorrect.
2. Statement I is incorrect and Statement II is correct.
3. Statement I is correct and Statement II is incorrect.
4. Both Statement I and Statement II are correct.
Subtopic:  Angular SHM |
Level 3: 35%-60%
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The time period of a simple pendulum in a stationary lift is \(T.\) If the lift accelerates upward with an acceleration of \(\dfrac g 6\) (where \(g\) is the acceleration due to gravity), then the time period of the pendulum would be:
1. \(\sqrt{\dfrac{6}{5}} ~T \) 2. \(\sqrt{\dfrac{5}{6}} ~T\)
3. \(\sqrt{\dfrac{6}{7}}~T\) 4. \(\sqrt{\dfrac{7}{6}} ~T\)
Subtopic:  Angular SHM |
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Level 1: 80%+
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The motion of a simple pendulum executing simple harmonic motion is represented by the equation;
\({y}={A} \sin (\pi {t}+\phi),\) where time is measured in seconds. The length of the pendulum is:
1. \(97.23~\text{cm}\) 
2. \(25.3~\text{cm}\) 
3. \(99.4~\text{cm}\) 
4. \(406.1~\text{cm}\) 
Subtopic:  Angular SHM |
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Level 1: 80%+
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The length of a seconds pendulum at a height \(h=2R\) from the earth's surface will be: 
(Given \(R=\) The radius of the earth and acceleration due to gravity at the surface of the earth \(g = \pi^2 ~\text{m/s}^2\))
1. \({\dfrac 2 9} ~\text m\) 2. \({\dfrac 4 9} ~\text m\)
3. \({\dfrac 8 9} ~\text m\) 4. \({\dfrac 1 9} ~\text m\)
Subtopic:  Angular SHM |
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Level 2: 60%+
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The time period of oscillation of a simple pendulum of length \(L,\) suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination \(\alpha,\) is given by:
1. \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{cos}\alpha}}\) 2. \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{sin}\alpha}}\)
3. \(2\pi \sqrt{\dfrac{L}{g}}\) 4. \(2\pi \sqrt{\dfrac{L}{g~ \mathrm{tan}\alpha}}\)
Subtopic:  Angular SHM |
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Level 3: 35%-60%
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