Identify correct statement among the following
Easy
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
- 1View SolutionIn forced oscillations, a particle oscillates simple harmonically with a frequency equal to
- 2A particle executes simple harmonic oscillation with an amplitude $a.$ The period of oscillation is $T.$ The minimum time taken by the particle to travel half of the amplitude from the equilibrium position isView Solution
- 3The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)View Solution
- 4The mass of a particle is $1\,\,kg$ and it is moving along $x-$ axis. The period of its small oscillation is $\frac {\pi }{2}$ . Its potential energy may beView Solution
- 5A mass $0.9\,kg$, attached to a horizontal spring, executes $SHM$ with an amplitude $A _{1}$. When this mass passes through its mean position, then a smaller mass of $124\,g$ is placed over it and both masses move together with amplitude $A _{2}$. If the ratio $\frac{ A _{1}}{ A _{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be$......$View Solution
- 6To find the spring constant $(k)$ of a spring experimentally, a student commits $2 \%$ positive error in the measurement of time and $1 \%$ negative error in measurement of mass. The percentage error in determining value of $\mathrm{k}$ is :View Solution
- 7The displacement equations of two interfering waves are given byView Solution
$y_1 =10 \sin \left(\omega t+\frac{\pi}{3}\right) cm$
$y_2 =5[\sin (\omega t)+\sqrt{3} \cos \omega t] \;cm$ respectively.
The amplitude of the resultant wave is $.............cm$.
- 8A particle moves on $x-$ axis according to the equation $x = x_0\,\,sin^2\,\omega t,$ the motion is simple harmonicView Solution
- 9View SolutionThe total energy of a particle executing S.H.M. is proportional to
- 10The 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