The mass of a mass-spring system is displaced and released from rest. If the 20 gm mass is observed to return to the release point every 2 seconds, determine the stiffness of the spring.

Short Answer

Expert verified
Answer: The stiffness of the spring in the mass-spring system is approximately 9.87 N/m.

Step by step solution

01

Convert given mass to kilograms

The given mass is 20 grams, but usually, when working with equations in physics, we'll need to use the mass unit in kilograms, so we need to convert this: 1 kg = 1000 grams m = 20 gm * (1 kg / 1000 gm) = 0.02 kg
02

Find the period T

The period is given as the time it takes for the mass to return to its initial position. In this case, T = 2 seconds.
03

Use the formula for the period of oscillation

The formula for the period of oscillation (T) in a mass-spring system is given by: T = 2 * pi * sqrt(m / k) Where "m" is the mass of the object, "k" is the spring constant, and "pi" is the mathematical constant approximately equal to 3.14159. In this case, we have T = 2 seconds and m = 0.02 kg. We want to solve for k.
04

Rearrange the formula to solve for k

We can rearrange the formula for the period of oscillation to solve for k: k = m / (T / (2 * pi))^2
05

Plug in the values and compute k

Now we can plug in the values we know (T = 2 seconds and m = 0.02 kg) and calculate k: k = 0.02 kg / (2 s / (2 * 3.14159))^2 k = 0.02 kg / (1 / 3.14159)^2 k ≈ 9.87 N/m So the stiffness of the spring is approximately 9.87 N/m.

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