Question:
Two simple pendulums of length 1 m and 4 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:
Two simple pendulums of length 1 m and 4 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:
Show Hint
\(T = 2\pi\sqrt{L/g}\); they are in phase again when the shorter one has gained exactly one full oscillation on the longer one.
Updated On: Oct 31, 2026
- 2
- 7
- 5
- 3
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
\(T \propto \sqrt{L}\), so \(\frac{T_2}{T_1} = \sqrt{\frac41} = 2\), i.e. \(T_2 = 2T_1\).
They are again in phase when the shorter pendulum has made one more complete oscillation than the longer one: if the shorter one makes \(n\) oscillations, the longer makes \(n-1\).
\(nT_1 = (n-1)T_2 = 2(n-1)T_1 \Rightarrow n = 2n - 2 \Rightarrow n = 2\).
Answer: (1) 2
They are again in phase when the shorter pendulum has made one more complete oscillation than the longer one: if the shorter one makes \(n\) oscillations, the longer makes \(n-1\).
\(nT_1 = (n-1)T_2 = 2(n-1)T_1 \Rightarrow n = 2n - 2 \Rightarrow n = 2\).
Answer: (1) 2
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ 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 centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main simple harmonic motion Questions
- Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is _____ degree.
- A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $\theta$ of thread deflection in the extreme position will be :
- When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is \(\frac{x}{ 8}\) , where x = _________.
- A particle performing simple harmonic motion is such that its amplitude is 4 m and speed of particle at mean position is 10 m/s. Find the distance of particle from mean position where velocity becomes 5 m/s.
- A particle is executing simple harmonic motion along the \(x - axis\), with amplitude \(A\), about the origin. Then the ratio of Kinetic energy and total energy at \(x = \frac{A}{3}\) is
View More Questions
Top JEE Main Questions
- Two masses $m$ 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 centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions