Frequency of oscillation

A weight of 1 kg was suspended from the spring, and the spring was extended by 1.5 cm. Determine the frequency of the natural oscillation of the resulting oscillator.

Correct answer:

f =  4.0701 Hz

Step-by-step explanation:

$$\begin{gathered} m=1\text{kg} \\ d=1.5\text{cm}\to \text{m}=1.5:100\text{m}=0.015\text{m} \\ g=9.81{\text{m/s}}^2 \\ F=m\cdot g=1\cdot 9.81=\frac{981}{100}=9.81\text{N} \\ F=k\cdot d \\ k=F/d=\frac{981}{100}/\frac{3}{200}=\frac{981}{100}:\frac{3}{200}=\frac{981}{100}\cdot \frac{200}{3}=\frac{981\cdot 200}{100\cdot 3}=\frac{196200}{300}=654 \\ f=\frac{1}{2\pi }\cdot \sqrt{\frac{k}{m}}=\frac{1}{2\cdot 3.1416}\cdot \sqrt{\frac{654}{1}}=4.0701\text{Hz} \end{gathered}$$

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