A 32.0-kg penguin on frictionless ice holds onto the end of a horizontal spring with spring constant 250 N/m and oscillates back and forth. What is its period of oscillation?
You must have come across the formula
periodn = 2 pi sqrt(M/k)
Use it.
Actually I didn't know that formula haha. Thanks!
To determine the period of oscillation of the penguin on the spring, we can use the equation:
T = 2π√(m/k)
where:
T is the period of oscillation,
m is the mass of the penguin, and
k is the spring constant.
Given:
m = 32.0 kg
k = 250 N/m
Plugging these values into the equation, we have:
T = 2π√(32.0 kg / 250 N/m)
Now, let's solve for T using the following steps:
Step 1: Calculate the square root of the mass over the spring constant.
√(32.0 kg / 250 N/m) = √(0.128 kg/N) = 0.358 kg/N
Step 2: Multiply the square root by π (pi).
0.358 kg/N x π ≈ 1.12 kg/N
Step 3: Multiply the result by 2.
1.12 kg/N x 2 = 2.24 kg/N
Therefore, the period of oscillation of the penguin on the spring is approximately 2.24 seconds.