What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10cm$ ......... $m/s$ (Take $g = 9.8\,m/{s^2})$
Easy
Download our app for free and get started
(b)$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 0.1} = \sqrt {1.96} = 1.4\,m/s$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1The potential energy of a particle of mass $0.1\,kg,$ moving along $x-$ axis, is given by $U = 5x(x-4)\,J$ where $x$ is in metres. It can be concluded thatView Solution
- 2Two simple harmonic motions are represented by equations ${y_1} = 4\,\sin \,\left( {10t + \phi } \right)$ and ${y_2} = 5\,\cos \,10\,t$ What is the phase difference between their velocities?View Solution
- 3View SolutionA simple pendulum is taken from the equator to the pole. Its period
- 4A particle executes simple harmonic motion with an amplitude of $4 \,cm$. At the mean position the velocity of the particle is $10\, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \,cm/s$ isView Solution
- 5View SolutionIdentify the function which represents a nonperiodic motion.
- 6A man weighing $60\, kg$ stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude $0.1\, m$ and frequency $\frac{2}{\pi }Hz$. Which of the following statement is correctView Solution
- 7The metallic bob of simple pendulum has the relative density $5$. The time period of this pendulum is $10\,s$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{ x } s$. The value of $x$ will be.View Solution
- 8The equation of a particle executing simple harmonic motion is given by $x =\sin \pi\left( t +\frac{1}{3}\right) m$. At $t =1 \,s$, the speed of particle will be .......... $cm s ^{-1}$ (Given : $\pi=3.14$ )View Solution
- 9A block of mass $200\, g$ executing $SHM$ under the influence of a spring of spring constant $K=90\, N\,m^{-1}$ and a damping constant $b=40\, g\,s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $ln\,\frac{1}{2} = -0.693$)View Solution
- 10The displacement of a particle along the $x-$ axis is given by $x=asin^2$$\omega t$ . The motion of the particle corresponds toView Solution