A mass is stretched 4cm from equilibrium position of a spring. If the displacement is doubled, what can be concluded about the restoring force?
A. It halved.
B. Same
C. Increased by 4
D. Decreased by 4
If the displacement of a mass attached to a spring is doubled, the restoring force will also be doubled. This means that the correct answer is B. The restoring force remains the same.
To determine the relationship between the displacement and the restoring force of a spring, we can refer to Hooke's Law, which states that the restoring force is directly proportional to the displacement from the equilibrium position.
Mathematically, Hooke's Law can be represented as follows:
F = -kx
Where F represents the restoring force, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we know that the mass is initially stretched 4 cm from the equilibrium position. Let's assume this displacement is represented by x1.
Next, we are told that the displacement is doubled. Let's denote this new displacement as 2x1 (since it's twice the initial displacement).
Now, we need to compare the restoring forces at the two displacements. Let's calculate the restoring force for each using Hooke's Law.
For the initial displacement (x1):
F1 = -kx1
For the doubled displacement (2x1):
F2 = -k(2x1)
Now, let's compare F2 to F1 to determine the relationship.
F2/F1 = (-k(2x1))/(-kx1)
The k cancels out:
F2/F1 = (2x1)/(x1)
F2/F1 = 2
From the equation, we can see that F2/F1 = 2, which means that the restoring force at the doubled displacement (F2) is twice the restoring force at the initial displacement (F1).
Therefore, the correct answer is B. The restoring force remains the same.
F = - k x
If you double the distance you double the force
4+4 = 8 increase by 4