A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ sec. If the lift accelerates upwards with an acceleration $\frac{g}{4}$, then the period of the pendulum will be
Medium
Download our app for free and get started
(c) In stationary lift $T = 2\pi \sqrt {\frac{l}{g}} $
In upward moving lift $T' = 2\pi \sqrt {\frac{l}{{(g + a)}}} $ ($a = $Acceleration of lift)
$ \Rightarrow \frac{{T'}}{T} = \sqrt {\frac{g}{{g + a}}} = \sqrt {\frac{g}{{\left( {g + \frac{g}{4}} \right)}}} = \sqrt {\frac{4}{5}} $
$ \Rightarrow T' = \frac{{2T}}{{\sqrt 5 }}$
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
- 1A particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$ then its amplitude will become ........ $cm$View Solution
- 2A simple perdulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will beView Solution
- 3The potential energy of a particle of mass $4\,kg$ in motion along the $x$-axis is given by $U =4(1-\cos 4 x )\,J$. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{ K }\right)\,s$. The value of $K$ is .......View Solution
- 4View SolutionA particle is moving with constant angular velocity along the circumference of a circle. Which of the following statements is true
- 5View SolutionWhen a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be.
- 6A pendulum with time period of $1\, s$ is losing energy due to damping. At certain time its energy is $45\, J$. If after completing $15\,oscillations$ , its energy has become $15\, J$, its damping constant (in $s^{-1}$ ) isView Solution
- 7View SolutionAn assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is :-
- 8View SolutionWhat is constant in S.H.M.
- 9The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \mathrm{~m}$. If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : [Use, $\mathrm{g}: 10 \mathrm{~ms}^{-2}$ ]View Solution
- 10Consider the following statements. The total energy of a particle executing simple harmonic motion depends on itsView Solution
$(1)$ Amplitude $(2) $ Period $(3)$ Displacement
Of these statements