A mass of 0.2,kg is attached to the lower end | vedclass

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

Question

(D) $1$. Equilibrium stretching: At equilibrium,the weight of the mass is balanced by the spring force. $mg = kx \implies 0.2 	imes 10 = 200 	imes x

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) $1$. Equilibrium stretching: At equilibrium,the weight of the mass is balanced by the spring force. $mg = kx \implies 0.2 	imes 10 = 200 	imes x \implies x = 0.01\,m = 1\,cm.$ Thus,statement $A$ is true.$2$. Motion from unstretched state: When the mass is raised to the unstretched position and released,the amplitude of oscillation $A$ is equal to the equilibrium extension $x = 1\,cm.$ The mass will oscillate between the unstretched position $(0\,cm)$ and the lowest point $(2A = 2\,cm)$. Thus,it will go down by $2\,cm$ before moving upwards. Statement $B$ is true.$3$. Frequency of oscillation: The frequency $f$ is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{200}{0.2}} = \frac{1}{2\pi} \sqrt{1000} \approx \frac{31.62}{6.28} \approx 5.03\,Hz.$ This is nearly $5\,Hz.$ Statement $C$ is true.Since all statements are true,the correct option is $D$.

$A$ mass of $0.2\,kg$ is attached to the lower end of a massless spring of force-constant $200\,N/m,$ the upper end of which is fixed to a rigid support. Which of the following statements is/are true?

Similar Questions

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by $m$,the time period becomes $5T/3$. Then the ratio of $m/M$ is

If a body of mass $0.98\, kg$ is made to oscillate on a spring of force constant $4.84\, N/m$,the angular frequency of the body is ..... $rad/s$. (in $.22$)

$A$ mass $m$ is suspended by means of two coiled springs which have the same length in the unstretched condition,as shown in the figure. Their force constants are $k_1$ and $k_2$ respectively. When set into vertical vibrations,the period will be:

$A$ spring has a certain mass suspended from it and its period for vertical oscillation is $T$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now

Vedclass Test Series

Exam Paper Generator

Online Exam Module