Question

A& block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.

Answer 1

A block of mass $0.5\mathrm{kg}$ hanging from a vertical spring executes simple harmonic motion of amplitude $0.1\mathrm{m}$ and time period $0.314\mathrm{s}$ . Find the maximum force exerted by the spring on the block.

Solution.

1) Only spring.

2) A block hanging from a vertical spring at rest.

3) A block hanging from a vertical spring executes simple harmonic motion.

The time period of the simple harmonic motion:

$$T = 2 \pi \sqrt {\frac {m}{k}}.$$

The stiffness of the spring:

$$k = \frac {4 \pi^ {2} m}{T ^ {2}}.$$

Hooke's law (modul):

$$F = k x _ {0};$$ $$F = m g;$$ $$m g = k x _ {0};$$ $$x _ {0} = \frac {m g}{k};$$

From diagram the maximum force:

$$F _ {m a x} = k (x _ {0} + x _ {m});$$ $$F _ {m a x} = k \left(\frac {m g}{k} + x _ {m}\right);$$ $$F _ {m a x} = m g + k x _ {m}.$$ $$F _ {m a x} = m g + \frac {4 \pi^ {2} m}{T ^ {2}} x _ {m}.$$ $$F _ {m a x} = 0. 5 \cdot 9. 8 + \frac {4 \cdot 3 . 1 4 ^ {2} \cdot 0 . 5}{0 . 3 1 4 ^ {2}} 0. 1 = 2 4. 9 (N).$$

Answer: $F_{max}24.9N$.