The length of a simple pendulum is increased by $2\%$. Its time period will
AIPMT 1997, Easy
Download our app for free and get started
$l_{2}=1.02 l_{1} .$
Time period $(T)=2 \pi \times \sqrt{\frac{l}{g}} \propto \sqrt{l}$
Therefore $\frac{T_{2}}{T_{1}}=\sqrt{\frac{l_{2}}{l_{1}}}=\sqrt{\frac{1.02 l_{1}}{l_{1}}}=1.01$
Thus time period increased by $1 \%$.
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 executes simple harmonic motion with an amplitude of $4 \mathrm{~cm}$. At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{cm}$, where $\alpha=$____________.View Solution
- 2A wooden cube (density of wood $'d'$ ) of side $'l'$ floats in a liquid of density $'\rho '$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period $'T'$. Then, $'T'$ is equal toView Solution
- 3A simple harmonic motion is represented by $F(t) = 10\sin \,(20\,t + 0.5)$. The amplitude of the $S.H.M.$ is $a$ $=$ ....View Solution
- 4In the following questions, match column $-I$ with column $-II$ and choose the correct optionsView Solution
- 5A simple pendulum consisting of a light inextensible string of length $\ell$ attached to a heavy small bob of mass $m$ is at rest. The bob is imparted a horizontal impulsive force which gives it a speed of $\sqrt{4 g \ell}$. The speed of the bob at its highest point is ( $g$ is the accelaration due to gravity)View Solution
- 6Two simple pendulum first of bob mass $M_1$ and length $L_1$ second of bob mass $M_2$ and length $L_2$. $M_1 = M_2$ and $L_1 = 2L_2$. If these vibrational energy of both is same. Then which is correctView Solution
- 7Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2,$ the time period of pendulum will beView Solution
- 8A particle of mass $m$ is located in a one dimensional potential field where potential energy is given by : $V(x) = A(1 -cos\, px)$ ,View Solution
where $A$ and $p$ are constant.
The period of small oscillations of the particle is
- 9On a planet a freely falling body takes $2 \,sec$ when it is dropped from a height of $8 \,m$, the time period of simple pendulum of length $1\, m$ on that planet is ..... $\sec$View Solution
- 10The acceleration of a particle performing S.H.M. is at a distance of $3\; cm$ from the mean position is $ 12\,cm/sec^2 $. Its time period is ..... $\sec$View Solution