Question

A spring has a frequency of 5 Hz when a 20 N weight is suspended from it. How far would the spring stretch from equilibrium when a 50 N weight is suspended from it?

Answer 1

Question #58345, Physics / Mechanics | Relativity

A spring has a frequency of $5\mathrm{Hz}$ when a $20\mathrm{N}$ weight is suspended from it. How far would the spring stretch from equilibrium when a $50\mathrm{N}$ weight is suspended from it?

**Solution:**

According to Hooke's Law:

$$F = k X;$$

where $k$ is the spring stiffness;

$X$ is spring deformation;

$F$ is the force applied to the spring (50 N)

Therefore,

$$X = \frac {F}{k}$$

The frequency of a mass attached to a spring:

$$f = \frac {1}{2 \pi \sqrt {\frac {m}{k}}};$$ $$k = 4 m \pi^ {2} f ^ {2};$$ $$m = \frac {W}{g}; (W - \text{is the suspended weight of } 20 \mathrm{~N})$$ $$k = \frac {4 W \pi^ {2} f ^ {2}}{g};$$ $$X = \frac {F g}{4 W \pi^ {2} f ^ {2}}$$ $$X = \frac {5 0 \times 9 . 8}{4 \times 2 0 \times \pi^ {2} \times 5 ^ {2}} = 0. 0 2 5 \mathrm{~m}$$

Answer: 0.025 m or 25 mm

http://www.AssignmentExpert.com/