A simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$, the percentage increase in the time period of the pendulum of increased length is ..... $\%$
AIEEE 2003,AIIMS 2001, Medium
Download our app for free and get started
(a) If initial length ${l_1} = 100$ then ${l_2} = 121$
By using $T = 2\pi \sqrt {\frac{l}{g}} $
$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{l_1}}}{{{l_2}}}} $
Hence, $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{100}}{{121}}} $
$\Rightarrow {T_2} = 1.1\,{T_1}$
$\%$ increase = $\frac{{{T_2} - {T_1}}}{{{T_1}}} \times 100 = 10\,\% $
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 SolutionThe diagram shows two oscillations. What is the phase difference between the oscillations?
- 2For a simple pendulum the graph between $L$ and $T$ will be.View Solution
- 3A particle is executing the motion $x = A\cos (\omega \,t - \theta )$. The maximum velocity of the particle isView Solution
- 4The function $sin^2\,(\omega t)$ representsView Solution
- 5A light balloon filled with helium of density $\rho_{ He }$ is tied to a long light string of length $l$ and the string is attached to the ground. If the balloon is displaced slightly in the horizontal direction from the equilibrium and released. Then,View Solution
- 6A spring mass system executes damped harmonic oscillations given by the equationView Solution
$y = A{e^{ - \frac{{bt}}{{2m}}}}\sin (\omega 't + \phi )$
where the symbols have their usual meanings. If a $2\ kg$ mass $(m)$ is attached to a spring of force constant $(K)$ $1250\ N/m$ , the period of the oscillation is $\left( {\pi /12} \right)s$ . The damping constant $‘b’$ has the value. ..... $kg/s$
- 7Which of the following quantities are always negative in a $SHM$View Solution
- 8Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T,$ respectively. At time $t=0$ one particle has displacement $A$ while the other one has displacement $\frac {-A}{2}$ and they are moving towards each other. If they cross each other at time $t,$ then $t$ isView Solution
- 9View SolutionA flat horizontal board moves up and down in $SHM$ of amplitude $\alpha$. Then the shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is
- 10When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude $a$ isView Solution