A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is removed and replaced by a 1.50-kg object, what will be the period of oscillation?
A)1.55 s
B)2.58 s
C)2.12 s
D)3.38 s
E)4.49 s

Question

Quizplus · Accepted Answer

The period of oscillation for an object hanging from a vertical spring is given by the formula: T = 2π√(m/k), where m is the mass of the object and k is the spring constant.

Since the spring is the same in both cases, the only variable that changes is the mass.

For the first case, T1 = 1.50 s with m1 = 0.750 kg.

For the second case, we have m2 = 1.50 kg.

Using the formula, we can solve for T2:

T2 = 2π√(m2/k)

We can find k by using the period of the first case:

T1 = 2π√(m1/k)

Solving for k:

k = (4π^2m1)/T1^2

Plugging in the values:

k = (4π^2 × 0.750 kg) / (1.50 s)^2

k = 10π^2 N/m

Now we can find T2:

T2 = 2π√(m2/k)

Plugging in the values:

T2 = 2π√(1.50 kg / (10π^2 N/m))

T2 ≈ 2.12 s

Therefore, the answer is C.

A 0750-Kg Object Hanging from a Vertical Spring Is Observed to to Oscillate