Question

A spring has natural length of $50\text{cm}$ and a force constant of $2.0\times {10}^3{\text{Nm}}^{-1}$ . A body of mass $10\text{kg}$ is suspended from it in equilibrium position. If the body is pulled down, such that length of spring becomes $58\text{cm}$ and released. If it executes simple harmonic motion, what is the net force on the body, when it is at it's lowermost position of oscillation ?
$(\text{Take g}=10{\text{ms}}^{-2})$.

• A. $20\text{N}$
• B. $40\text{N}$
• C. $60\text{N}$
• D. $80\text{N}$

Answer 1

The correct option is C $60\text{N}$

Given that,
Natural length of the spring $l_n=50\text{cm}$
Force constant of the spring $k=2.0\times {10}^3{\text{Nm}}^{-1}$
Extension in the length of the spring when $10\text{kg}$ mass is suspended.
$kx=mg$
$\Rightarrow x=\frac{10\text{kg}\times 10{\text{ms}}^{-2}}{2.0\times {10}^3{\text{Nm}}^{-1}}$
$\Rightarrow x=0.05\text{m}=5\text{cm}$
Equilibrium position of the mass is
$x_1=50\text{cm}+5\text{cm}$
$x_1=55\text{cm}$
Now spring is further stretched to $58\text{cm}$
So, displacement from mean position is
$x^{\prime }=58\text{cm}-55\text{cm}=3\text{cm}$

Net force at lowermost position is,
$F_{net}=k{x}^{\prime }=2\times {10}^3\times 3\times {10}^{-2}=60\text{N}$
Hence, ${}^{\prime }{c}^{\prime }$ is the correct option.