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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the spring constant of the spring?

A) 24.1 N/m
B) 2.45 N/m
C) 0.102 N/m
D) 12.1 N/m
E) 0.610 N/m

Answer :

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

m is the mass of object

[tex]m=\dfrac{W}{g}[/tex]

[tex]m=\dfrac{2.45}{9.8}[/tex]

m = 0.25 kg

[tex]k=\dfrac{4\pi^2m}{T^2}[/tex]

[tex]k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}[/tex]

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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