Question

a spring is stretched by 5 cm when a mass of 70 gm is hung on it. If total mass of 100 gm is hung on spring and the mass starts vertical oscillation. what is the spring constant k ?

Answer 1

To solve the problem it is necessary to apply the Hooke Law, which is given by.

$F_{K}=K\;X$

Where.

• The deformation is $X=5 cm \times \frac{1\;m}{100\;cm}=0.05\;m$
• The constant is $K$

The restorative force is equal to the mass of the intended placement.

$F_{k}=W$

The weight placed is

$W=m\;g$

Where.

• The mass is $m=70 g\times \frac{1\;kg}{1000\;g}=0.07\;Kg$
• The acceleration of gravity is $g=9.8\;\frac{m}{s^{2}}$

Evaluating numerically.

$W=0.07\;Kg\times 9.8\frac{m}{s^{2}}\\ W=0.686\;N$

The expression for the elastic constant is

$F_{k}=K\;X \\ K\;X=F_{K} \\ K=\frac{F_{k}}{X}$

Numerically evaluating to calculate the constant.

$K=\frac{0.686\;N}{0.05\;m} =13.72\;\frac{N}{m}$

The spring constant is $\boxed{K=13.72\;\frac{N}{m}}$