Three masses 100 text{ g}, 300 text{ g}, and | vedclass

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

Question

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.Initially,

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.Initially, the masses are $m_1 = 100 	ext{ g}$, $m_2 = 300 	ext{ g}$, and $m_3 = 500 	ext{ g}$.When the $500 	ext{ g}$ mass is removed, the remaining mass is $M_1 = 100 	ext{ g} + 300 	ext{ g} = 400 	ext{ g}$.The period is $T_1 = 3 	ext{ s}$.So, $3 = 2\pi \sqrt{\frac{400}{k}}$.When the $300 	ext{ g}$ mass is also removed, the remaining mass is $M_2 = 100 	ext{ g}$.Let the new period be $T_2$.So, $T_2 = 2\pi \sqrt{\frac{100}{k}}$.Dividing the two equations:$\frac{T_2}{3} = \frac{2\pi \sqrt{100/k}}{2\pi \sqrt{400/k}} = \sqrt{\frac{100}{400}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore, $T_2 = 3 	imes \frac{1}{2} = 1.5 	ext{ s}$.

Similar Questions

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$,where the value of $\alpha$ is . . . . . . .

$A$ wooden block performs $SHM$ on a frictionless surface with frequency $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown,then the $SHM$ of the block will be

The springs in the figure are identical,but the length of the spring in $A$ is three times that in $B$. The ratio of the time periods $T_A/T_B$ is:

The equivalent spring constant of the system as shown in the figure will be:

Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module