What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mine
Easy
Download our app for free and get started
(b) In deep mine $g' = g\,\left( {1 - \frac{d}{R}} \right)$; i.e., $g$ decreases so according to $n \propto \sqrt g ,$ frequency also decreases.
Download our app
and 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.*
The displacement time equation of a particle executing $SHM$ is : $x = A \,sin\,(\omega t + \phi )$. At time $t = 0$ position of the particle is $x = A/2$ and it is moving along negative $x-$ direction. Then the angle $\phi $ can be
A plank with a small block on top of it is under going vertical $SHM.$ Its period is $2\, sec.$ The minimum amplitude at which the block will separate from plank is :
A block of mass $2\,kg$ is attached with two identical springs of spring constant $20\,N / m$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in SI unit. The value of $x$ is $..........$
Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0,$ one particle is at displacement $y_1= +A$ and the other at $y_2= -A/2,$ and they are approaching towards each other. After what time they cross each other ?
A particle is performing simple harmonic motion along $x-$axis with amplitude $4 \,cm$ and time period $1.2\, sec$. The minimum time taken by the particle to move from $x =2 ,cm$ to $ x = + 4\, cm$ and back again is given by .... $\sec$
A body of mass $m $ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released , it oscillates with a time period of $3\,s$ . When the mass $m$ is increased by $1\,kg$ , the time period of oscillations becomes $5\,s$ . The value of $m$ in $kg$ is
If a simple harmonic oscillator has got a displacement of $0.02\, m$ and acceleration equal to $2.0\,m{s^{ - 2}}$ at any time, the angular frequency of the oscillator is equal to .... $rad\,{s^{ - 1}}$