In case of a simple pendulum, time period versus length is depicted by
A
B
C
D
Easy
Download our app for free and get started
B
b (b)$T \propto \sqrt l \Rightarrow {T^2} \propto l$
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.*
A simple pendulum is attached to a block which slides without friction down an inclined plane $A B C$ having an angle of inclination $\alpha$ as shown below. While the block is sliding down the pendulum oscillates in such a way that at its mean position the direction of the string is
A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then, its time period of vibration will be
The graph shown was obtained from experimental measurements of the period of oscillations $T$ for different masses $M$ placed in the scale pan on the lower end of the spring balance. The most likely reason for the line not passing through the origin is that the
Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega $ along the $x-$ axis. Their mean position is separated by $X_0$ (where $X_0 > A$ ). If maximum separation between them is $X_0 + 2A$ , then the phase difference between their motion is
A block of rectangular size of mass $m$ and area of cross-section $A$, floats in a liquid of density $\rho$. If we give a small vertical displacement from equilibrium, it undergoes $S H M$ with time period $T$, then