Question

A spring is stretched by 5 cm by a force of 10 N. The time period of the oscillations, when a mass of 2 kg is suspended by it, is

• A. 0.628 s
• B. 0.0628 s
• C. 6.28 s
• D. 3.14 s

Answer 1

The Correct Option is A

To find the time period of oscillation of the system, we need to use the formula for the time period of a mass-spring system:

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

where \(T\) is the time period, \(m\) is the mass, and \(k\) is the spring constant. First, we need to determine the spring constant \(k\) using Hooke's Law:

\(F = k \cdot x\)

where \(F\) is the force applied and \(x\) is the displacement.

Given: \(F = 10\, \text{N}\) and the spring is stretched by \(x = 5\, \text{cm} = 0.05\, \text{m}\)

Plug these values into Hooke's Law to find \(k\):

\(10 = k \cdot 0.05\)

Solving for \(k\) gives:

\(k = \frac{10}{0.05} = 200\, \text{N/m}\)

Now, use the formula for \(T\) with \(m = 2\, \text{kg}\) and \(k = 200\, \text{N/m}\):

\(T = 2\pi \sqrt{\frac{2}{200}}\)

\(T = 2\pi \sqrt{0.01}\)

\(T = 2\pi \times 0.1\)

\(T = 0.2\pi\)

Using the approximation \(\pi \approx 3.14\):

\(T \approx 0.2 \times 3.14 = 0.628\, \text{s}\)

Therefore, the correct answer is \(0.628\, \text{s}\).

Thus, the time period of the oscillations when a mass of 2 kg is suspended by it is 0.628 s.