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Q68P

Expert-verifiedFound in: Page 440

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A 2.00 kg**** block hangs from a spring. A 300 kg**** body hung below the block stretches the spring 2.00 cm**** farther. **

**What is the spring constant?****If the 300 kg****body is removed and the block is set into oscillation, find the period of the motion.**

a) The spring constant is 147 N/m .

b) The period of motion is 0.733 s .

- Mass of the block,m=2.00 kg .
- Mass of the body,${m}_{b}=300\mathrm{gm}\mathrm{or}0.30\mathrm{kg}$ .
- The extension of the spring,x=2.00 cm or 0.02 m .

**Hooke’s law states that the restoring force is directly proportional to the displacement of the oscillating body and acts in the opposite direction to the displacement. The extra weight (body) attached to the string stretches the spring. The spring obeys Hooke’s law and its motion exhibits simple harmonic motion.**

** **

Formula:

The stretched force applied on a body,$F=-kx$ (i)

The period of oscillations, $T=\frac{2\tau \tau}{\omega}$ (ii)

The angular frequency of an oscillation, $\omega =\sqrt{\frac{k}{m}}$ (iii)

The body attached to the block stretches the spring by amount x. The spring obeys Hooke’s law. Hence, we can write,

$F=\mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{body}\phantom{\rule{0ex}{0ex}}={\mathrm{m}}_{\mathrm{b}}\mathrm{g}\phantom{\rule{0ex}{0ex}}=0.30\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}$

So, considering the magnitude only using equation (i), we get the spring constant as:$k=\frac{F}{x}\phantom{\rule{0ex}{0ex}}=\frac{0.30\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}}{0.02\mathrm{m}}\phantom{\rule{0ex}{0ex}}=147\mathrm{N}/\mathrm{m}$

Hence, the value of spring constant is 147 N/m .

Using equations (ii) and (iii), we get the period of oscillations as:

$T=2\tau \tau \sqrt{\frac{m}{k}}\phantom{\rule{0ex}{0ex}}=2\times 3.14\times \sqrt{\frac{2.00kg}{147N/m}}\phantom{\rule{0ex}{0ex}}=0.733s$

Hence, the value of period is 0.733 s .

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