A body executes simple harmonic motion under | vedclass

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

Question

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$, the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = kx$, we have $k = \fra

Vedclass · Accepted Answer

$\frac{12}{25}$

Vedclass · Answer

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$, the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = kx$, we have $k = \frac{F}{x}$, so $T = 2\pi \sqrt{\frac{mx}{F}}$.This implies $T^2 \propto \frac{1}{F}$, or $F \propto \frac{1}{T^2}$.Let $F_1 = \frac{c}{T_1^2}$ and $F_2 = \frac{c}{T_2^2}$, where $c$ is a constant.When both forces act simultaneously in the same direction, the net force is $F_{net} = F_1 + F_2$.The new time period $T$ satisfies $F_{net} = \frac{c}{T^2}$.Substituting the expressions: $\frac{c}{T^2} = \frac{c}{T_1^2} + \frac{c}{T_2^2}$.$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}} = \frac{(4/5) 	imes (3/5)}{\sqrt{(4/5)^2 + (3/5)^2}} = \frac{12/25}{\sqrt{16/25 + 9/25}} = \frac{12/25}{\sqrt{25/25}} = \frac{12}{25} \ s$.

Similar Questions

$A$ particle executes $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When the displacement of the particle from the mean position is $1 \,cm$,the magnitude of its acceleration is equal to the magnitude of its velocity. The time period of oscillation is ........

$A$ uniform rod of mass $m$ and length $l$ is suspended about its end. The time period of small angular oscillations is ..........

$A$ sphere of radius '$r$' is kept on a concave mirror of radius of curvature '$R$'. The arrangement is kept on a horizontal table. If the sphere is displaced from its equilibrium position and left,then it executes $S$.$H$.$M$. The period of oscillation will be ($g =$ acceleration due to gravity).

$A$ man of mass $M$ is swinging on a swing made of $2$ ropes of equal length $L$. The distance between the points of suspension is also $L$. The time period of the small oscillations about the mean position in a direction perpendicular to the plane of the swing is:

$A$ ring is hung on a nail. It can oscillate,without slipping or sliding $(i)$ in its plane with a time period $T_{1}$ and,$(ii)$ back and forth in a direction perpendicular to its plane,with a period $T_{2}$. The ratio $\frac{T_{1}}{T_{2}}$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module