When a body of mass 1.0 ~kg is suspended from | vedclass

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

Question

(B) Given: Mass $m_1 = 1.0 ~kg$,extension $l_1 = 5 ~cm = 0.05 ~m$.Using Hooke's law,$m_1 g = k l_1$,where $k$ is the spring constant.$k = \frac{m_1 g}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given: Mass $m_1 = 1.0 ~kg$,extension $l_1 = 5 ~cm = 0.05 ~m$.Using Hooke's law,$m_1 g = k l_1$,where $k$ is the spring constant.$k = \frac{m_1 g}{l_1} = \frac{1.0 	imes 10}{0.05} = 200 ~N/m$.Now,for a mass $m_2 = 2.0 ~kg$ suspended from the same spring,the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$.$\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 ~rad/s$.The block is pulled by $A = 10 ~cm = 0.1 ~m$,which represents the amplitude of the simple harmonic motion.The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$.$v_{\max} = 0.1 ~m 	imes 10 ~rad/s = 1 ~m/s$.

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

When a body of mass $1.0 ~kg$ is suspended from a certain light spring hanging vertically,its length increases by $5 ~cm$. By suspending a $2.0 ~kg$ block to the spring,if the block is pulled through $10 ~cm$ and released,the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 ~m/s^2$)

Similar Questions

The locus of the midpoints of the chords of the parabola $y^2 = 4ax$ which pass through the origin is:

The four distinct points $(0,0), (2,0), (0,-2)$ and $(k,-2)$ are concyclic,if $k$ is equal to

When a cell is fully turgid,which of the following will be zero?

Which one of the following is a correct set?

Vedclass Test Series

Exam Paper Generator

Online Exam Module