Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is 7 : 8, then the ratio of

Q: Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$, then the ratio of lengths of the two pendulums will be

d (d) Suppose at $t = 0$, pendulums begins to swing simultaneously. Hence, they will again swing simultaneously if ${n_1}{T_1} = {n_2}{T_2}$ $ \Rightarrow \frac{{{n_1}}}{{{n_2}}} = \frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}}$ $\Rightarrow \frac{{{l_1}}}{{{l_2}}} = {\left( {\frac{{{n_2}}}{{{n_1}}}} \right)^2} = {\left( {\frac{8}{7}} \right)^2} = \frac{{64}}{{49}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$, then the ratio of lengths of the two pendulums will be

A$7:8$
B$8:7$
C$49 : 64$
D$64:49$

Easy

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

Download our app
and 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.*

Download App

Similar Questions

1
A particle of mass $1\, {kg}$ is hanging from a spring of force constant $100\, {Nm}^{-1 .}$ The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period ${T}$. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{{T}}{{x}}$. The value of ${x}$ is ..... .
View Solution
2
A particle starts from a point $P$ at a distance of $A/2$ from the mean position $O\, \&$ travels towards left as shown in the figure. If the time period of $SHM,$ executed about $O$ is $T$ and amplitude $A$ then the equation of motion of particle is :
View Solution
3
Two massless springs with spring constants $2\,k$ and $2\,k$, carry $50\, g$ and $100 \,g$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be
View Solution
4
A body performs $S.H.M.$ Its kinetic energy $K$ varies with time $t$ as indicated by graph
View Solution
5
If a Second's pendulum is moved to a planet where acceleration due to gravity is $4$ times, the length of the second's pendulum on the planet should be made .......... times
View Solution
6
The motion of a simple pendulum excuting $S.H.M$. is represented by following equation.

$Y = A \sin (\pi t +\phi)$, where time is measured in $second$.

The length of pendulum is .............$cm$
View Solution
7
The variation of kinetic energy $(KE)$ of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position $(A)$ is given by
View Solution
8
A pendulum of length $2\,m$ lift at $P$. When it reaches $Q$, it losses $10\%$ of its total energy due to air resistance. The velocity at $Q$ is .... $m/sec$
View Solution
9
In the reported figure, two bodies $A$ and $B$ of masses $200\, {g}$ and $800\, {g}$ are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be $.....\,{rad} / {s}$ when ${k}=20 \,{N} / {m} .$
View Solution
10
A simple pendulum executing $S.H.M.$ is falling freely along with the support. Then
View Solution

Download our appand get started for free

Similar Questions

Download our app
and get started for free