Simple Harmonic Motion of Vertical Spring

Question

Suppose a spring stretches 3.9 cm when an object with mass 10 g is hung from it. Now, the object is replaced with a block with mass 25 g. The mass oscillates in simple harmonic motion. Find the period of the block's motion.

Answer 1

Calculate the spring constant from the initial data given:

1. Draw a free-body diagram:

1. ∑F_y = 0 and ∑F_y are F_s - mg so F_s - mg = 0 or F_s = mg
2. Plug in m = 10/1000 kg and g = 9.8 m/s² to get F_s = 0.098 N
3. Use Hooke's Law to find the spring constant "k"
4. F_s = kx so k = F_s/x = 0,098/(3.9/100) = 2.513 N/m

Calculate the period using T = 2π√(m/k)

1. Plug in k = 2.513 and m = 25/1000
2. T = 2π√(0.025/2.513) = 2π√0.00995 = 2π(0.0997) = 0.63 seconds

Answer 2

Answer 3

Step 1: Use Hooke’s Law to find the spring constant

Hooke's Law:

F = kx

is the force due to the hanging mass (10 g = 0.010 kg)

k = \frac{mg}{x} = \frac{0.010 \cdot 9.8}{0.039} \approx 2.51 \, \text{N/m}

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Step 2: Use the formula for the period of a mass-spring system

T = 2\pi \sqrt{\frac{m}{k}}

T = 2\pi \sqrt{\frac{0.025}{2.51}} \approx 2\pi \sqrt{0.00996} \approx 2\pi (0.0998) \approx 0.627 \, \text{s}

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Final Answer:

\boxed{T \approx 0.63 \, \text{seconds}}