The acceleration due to gravity on the surface of moon is 1.7 ms-2.What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms-2
The acceleration due to gravity on the surface of moon is 1.7 ms-2.What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms-2
Solution and Explanation
Acceleration due to gravity on the surface of moon, g'=1.7 ms-2
Acceleration due to gravity on the surface of earth, g=9.8 ms-2
Time period of a simple pendulum on earth, T=3.5 s
\(T=2π\sqrt\frac{l}{g}\)
Where,
l is the length of the pendulum
\(∴l=\frac{T^2}{(2π)^2}×g\)
\(=\frac{(3.5)^2}{4×(3.14)^2}×9.8 m\)
The length of the pendulum remains constant.
On moon’s surface, time period, \(T'=2π\sqrt\frac{l}{g'}\)
\(=2π\sqrt\frac{\frac{(3.5)^2}{4×(3.14)^2}×9.8}{1.7}=8.4 s\)
Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
Top Questions on Oscillations
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Physics
- Oscillations
- A spring is stretched by \( 0.2 \, \text{m} \) when a mass of \( 0.5 \, \text{kg} \) is suspended to it. The time period of the spring when \( 0.5 \, \text{kg} \) mass is replaced with a mass of \( 0.25 \, \text{kg} \) is (Acceleration due to gravity \( = 10 \, \text{m/s}^2 \))
- AP EAPCET - 2025
- Physics
- Oscillations
- The equations for the displacements of two particles in simple harmonic motion are \( y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right) \) and \( y_2 = 0.1\cos(\pi t) \) respectively. The phase difference between the velocities of the two particles at a time \( t = 0 \) is
- AP EAPCET - 2025
- Physics
- Oscillations
- If the function $\sin^2 \omega t$ (where $t$ is time in seconds) represents a periodic motion, then the period of the motion is
- AP EAPCET - 2025
- Physics
- Oscillations
- When the mass attached to a spring is increased from 4 kg to 9 kg, the time period of oscillation increases by \(0.2\pi\) s. Then the spring constant of the spring is
- AP EAPCET - 2025
- Physics
- Oscillations
Questions Asked in CBSE Class XI exam
- We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
- CBSE Class XI
- Motion in a straight line
- If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}- CBSE Class XI
- Complement of a Set
- What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH- CBSE Class XI
- Oxidation Number
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Write the following as intervals:
(i) {\(x: x ∈ R, -4 < x ≤ 6\)}
(ii) {\(x: x ∈ R, -12 < x < -10\)}
(iii) {\(x: x ∈ R, 0 ≤ x < 7\)}
(iv) {\(x: x ∈ R, 3 ≤ x ≤ 4\)}- CBSE Class XI
- Sets and their Representations