- 1(current)
Q. No.A uniform rod of length \(l\) is suspended by an end and is made to undergo small oscillations. The time period of small oscillation is \(T\). Then, the acceleration due to gravity at this place is:
| 1. | \(4\pi^2\dfrac{l}{T^2}\) | 2. | \(\dfrac{4\pi^2}{3}\dfrac{l}{T^2}\) |
| 3. | \(\dfrac{8\pi^2}{3}\dfrac{l}{T^2}\) | 4. | \(\dfrac{12\pi^2l}{T^2}\) |
Subtopic: Angular SHM |
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A physical pendulum consists of a uniform rod \(AB\) of mass \(m\) and length \(L,\) suspended from one end \(A\) – so as to rotate freely under gravity. If it is displaced slightly from its mean position, it executes SHM. Let the maximum kinetic energy of the rod be \(E_0.\)
If the time period of a simple pendulum of the same length is \(T_0,\) then the time period of this pendulum is:
If the time period of a simple pendulum of the same length is \(T_0,\) then the time period of this pendulum is:
| 1. | \({\Large\sqrt\frac23}T_0\) | 2. | \({\Large\sqrt\frac{1}{12}}T_0\) |
| 3. | \({\Large\sqrt\frac32}T_0\) | 4. | \(T_0\) |
Subtopic: Angular SHM |
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An elastic ball is projected vertically upward with a speed \(u,\) and it returns to the ground and rebounds, the motion is periodic with a period \(T.\) A simple pendulum, having a length equal to maximum altitude attained by this ball, would have a time period of:
| 1. | \(T\) | 2. | \(\pi T\) |
| 3. | \(\pi\sqrt2T\) | 4. | \(\dfrac{\pi}{\sqrt 2}T\) |
Subtopic: Angular SHM |
52%
From NCERT
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Three particles of equal mass are connected by massless rods of length \(L\) to form an equilateral triangle, \(ABC.\) This triangle is pivoted at \(A\) and allowed to oscillate in its own plane. The time period of small oscillation is:
| 1. | \(2\pi\sqrt{\dfrac{L}{g}}\) | 2. | \(2\pi\sqrt{\dfrac{2L}{g}}\) |
| 3. | \(2\pi\sqrt{\dfrac{L}{2g}}\) | 4. | \(2\pi\sqrt{\dfrac{2L}{\sqrt3g}}\) |
Subtopic: Angular SHM |
54%
From NCERT
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Two identical simple pendulums are compared, one \((A)\) located on the surface of the earth and the other \((B)\) – at a height \((h)\) above the earth's surface: \(h=\dfrac{R}{1000}.\)
Their time periods are related as:
1. \(T_A\Big(1+\dfrac{1}{1000}\Big)=T_B\)
2. \(T_B\Big(1+\dfrac{1}{1000}\Big)=T_A\)
3. \(T_A\Big(1+\dfrac{1}{2000}\Big)=T_B\)
4. \(T_B\Big(1+\dfrac{1}{2000}\Big)=T_A\)
Their time periods are related as:
1. \(T_A\Big(1+\dfrac{1}{1000}\Big)=T_B\)
2. \(T_B\Big(1+\dfrac{1}{1000}\Big)=T_A\)
3. \(T_A\Big(1+\dfrac{1}{2000}\Big)=T_B\)
4. \(T_B\Big(1+\dfrac{1}{2000}\Big)=T_A\)
Subtopic: Angular SHM |
60%
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A simple pendulum of time period \(T_0\) is taken in a rocket which is accelerating upwards initially and then, after some time, it moves with uniform velocity upward. The time period of the pendulum is observed within the rocket and is found to be \(2T_0\). The rocket, at that time, must be at a distance (above the earth's surface) of (radius of earth = \(R\))
| 1. | \(\dfrac{R}{2}\) | 2. | \(\dfrac{R}{4}\) |
| 3. | \(R\) | 4. | \(4R\) |
Subtopic: Angular SHM |
58%
From NCERT
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Given below are two statements:
| Assertion (A): | The time period of small oscillations of a simple pendulum is independent of the amplitude. |
| Reason (R): | The oscillatory motion of a simple pendulum is affected by the mass of the bob and the acceleration due to gravity, and so the time period depends only on these quantities. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
Subtopic: Angular SHM |
64%
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A simple pendulum undergoing small oscillations of amplitude \(1~\text{cm},\) has a time period of \(2~\text{s}.\) If the amplitude of oscillation is halved, the time period will:
| 1. | remain unchanged |
| 2. | increase to \(4~\text{s}\) |
| 3. | decrease to \(1~\text{s}\) |
| 4. | decrease to \(0.5~\text{s}\) |
Subtopic: Angular SHM |
74%
From NCERT
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A uniform rod \(AB\) of mass \(M\) is pivoted at its end \(A\) so that it can rotate about \(A\) in a vertical plane. The other end \(B\) is connected to a fixed vertical spring of spring constant \(k.\) If the rod were to perform small rotational oscillations about its horizontal mean position, the angular frequency \((\omega)\) would be:
| 1. | \(\sqrt{\Large\frac{k}{M}}\) | 2. | \(\sqrt{\Large\frac{12k}{M}}\) |
| 3. | \(\sqrt{\Large\frac{3k}{M}}\) | 4. | \(\sqrt{\Large\frac{3k}{2M}}\) |
Subtopic: Angular SHM |
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