A ball on a spring oscillates with a period of 4.14 s. What is the period of oscillation if the spring constant in the spring is quadrupled? 0.242 s 16.6 s 2.07 s 1.03 s
Question
A ball on a spring oscillates with a period of 4.14 s. What is the period of oscillation if the spring constant in the spring is quadrupled? 0.242 s 16.6 s 2.07 s 1.03 s
Solution
The period of oscillation of a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.
From the problem, we know that the initial period T1 = 4.14 s and the spring constant is quadrupled, meaning k2 = 4k1.
We want to find the new period T2 when the spring constant is quadrupled.
Since the mass m does not change, we can write the ratio of the two periods as T1/T2 = √(k2/k1).
Substituting the given values, we get T1/T2 = √(4k1/k1) = √4 = 2.
Therefore, T2 = T1/2 = 4.14 s / 2 = 2.07 s.
So, the period of oscillation if the spring constant in the spring is quadrupled is 2.07 s.
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