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A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed to a horizontal support. The student is asked to find the spring constant, k . After suspending a mass of 255.0 g from the spring, the student notices the spring is displaced 35.5 cm from its previous equilibrium. With this information, calculate the spring constant.

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Answer:

7.04661971831 N/m

Explanation:

x = Displacement of the spring = 35.5 cm

g = Acceleration due to gravity = [tex]9.81\ m/s^2[/tex]

m = Mass of block = 255 g

Here the forces are conserved

[tex]kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{0.255\times 9.81}{0.355}\\\Rightarrow k=7.04661971831\ N/m[/tex]

The spring constant is 7.04661971831 N/m

onyebuchinnaji

The spring constant of the spring with the given mass and displacement is 7.04 N/m.

The given parameters;

  • mass of the suspended, m = 255 g = 0.255 kg
  • displacement, x = 35.5 cm = 0.355 m

The spring constant of the spring is calculated by applying Hook's law as follows;

F = kx

mg = kx

[tex]k = \frac{mg}{x} \\\\k = \frac{0.255 \times 9.8}{0.355} \\\\k = 7.04 \ N/m[/tex]

Thus, the spring constant of the spring with the given mass and displacement is 7.04 N/m.

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