An assembly of identical spring-mass systems | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) If there were no collision,each spring-mass system would oscillate with a time period of $T_0 = 2 \pi \sqrt{\frac{m}{k}}$.Because of the elastic c

Vedclass · Accepted Answer

$\pi \sqrt{\frac{m}{k}}$

Vedclass · Answer

(C) If there were no collision,each spring-mass system would oscillate with a time period of $T_0 = 2 \pi \sqrt{\frac{m}{k}}$.Because of the elastic collision between the two masses,the masses are restricted from moving beyond their equilibrium positions towards each other. Each mass only performs half of its full oscillation cycle (the compression phase).Therefore,the time period of the oscillation of the system is half of the standard time period of a single spring-mass system.$T = \frac{T_0}{2} = \frac{1}{2} 	imes 2 \pi \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{m}{k}}$.

Similar Questions

With a resistance of ' X ' in the left gap and a resistance of $9 \Omega$ in the right gap of meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance $3 \Omega$ resistance be connected to obtained the balance at 50 cm from the left end?

$A$ diode detector is used to detect an amplitude-modulated wave of $60\%$ modulation by using a capacitor of capacity $250 \text{ pF}$ in parallel with a load resistance of $100 \text{ k}\Omega$. Find the maximum modulation frequency which could be detected by it.

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

Humoral immunity is due to

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module