If a particle is executing simple harmonic motion, then acceleration of particle
Easy
Download our app for free and get started
(c)
If a particle is executing S.H.M.
$a \alpha-\omega^2 x$
Hence it is not uniform and depends on $x$ rather than time.
Hence answer is $(c)$.
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.*
Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$
A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
A mass $m$ attached to a spring oscillates every $2\, sec$. If the mass is increased by $2 \,kg$, then time-period increases by $1\, sec$. The initial mass is ..... $kg$
A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.
The general displacement of a simple harmonic oscillator is $x = A \sin \omega t$. Let $T$ be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t=\frac{T}{\beta}$. The value of $\beta$ is $.........$
Two particle executing $S.H.M.$ of same amplitude of $20 \,cm$ with same period along the same line about same equilibrium position. The maximum distance between the two is $20 \,cm$. Their phase difference in radian is equal to