The potential energy of a particle of mass $100 \,g$ moving along the $x$-axis is given by $U = 5x(x - 4)$,where $x$ is in metres. The period of oscillation is .................

  • A
    $0.2 \pi \, s$
  • B
    $0.1 \pi \, s$
  • C
    $\pi \, s$
  • D
    $\frac{1}{\pi} \, s$

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Similar Questions

The potential energy of a particle of mass $4 \, kg$ in motion along the $x$-axis is given by $U = 4(1 - \cos 4x) \, J$. The time period of the particle for small oscillations $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) \, s$. The value of $K$ is .......

Two blocks of masses $m$ and $M$ $(M > m)$ are placed on a frictionless table as shown in the figure. $A$ massless spring with spring constant $k$ is attached to the lower block. If the system is slightly displaced and released,then ($\mu =$ coefficient of friction between the two blocks):
$(A)$ The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{M + m}{k}}$
$(B)$ The acceleration of the blocks is $a = \frac{kx}{M + m}$ ($x =$ displacement of the blocks from the mean position)
$(C)$ The magnitude of the frictional force on the upper block is $f = \frac{mkx}{M + m}$
$(D)$ The maximum amplitude of the upper block,if it does not slip,is $A = \frac{\mu mg(M + m)}{mk} = \frac{\mu g(M + m)}{k}$ (Wait,let's re-evaluate: $f_{max} = \mu mg$. Since $f = ma = m \cdot \frac{kx}{M+m}$,at max amplitude $A$,$m \cdot \frac{kA}{M+m} = \mu mg \implies A = \frac{\mu g(M+m)}{k}$)
$(E)$ Maximum frictional force can be $\mu mg$.
Choose the correct answer from the options given below.

Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and the same mass $m$ is suspended with them. If $T$ is the new time period in this position,then:

$A$ block of mass $m$ rests on a platform. The platform is given up and down $SHM$ with an amplitude $d$. What is the maximum frequency so that the block never leaves the platform?

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$,where the value of $\alpha$ is . . . . . . .

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