Question

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses \(1\,\text{kg}\) and \(0.2\,\text{kg}\) with a separation more than spring natural length and then released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is _______. (Given \(k=150\,\text{N/m}\)) 

• A. \(27\)
• B. \(30\)
• C. \(5\)
• D. \(20\)

Hint:

For two masses connected by a spring on a frictionless surface, always use reduced mass to find angular frequency.

Answer 1

The Correct Option is B

To find the angular frequency of the system, we use the formula for the angular frequency of a simple harmonic oscillator, which is given by:

\(\omega = \sqrt{\frac{k}{\mu}}\)

where \(\omega\) is the angular frequency, \(k\) is the spring constant, and \(\mu\) is the reduced mass of the system.

The reduced mass \(\mu\) for two masses \(m_1\) and \(m_2\) is calculated as:

\(\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\)

Given:

• \(m_1 = 1\,\text{kg}\)
• \(m_2 = 0.2\,\text{kg}\)
• \(k = 150\,\text{N/m}\)

First, calculate the reduced mass \(\mu\):

\(\mu = \frac{1 \cdot 0.2}{1 + 0.2} = \frac{0.2}{1.2} = \frac{1}{6}\,\text{kg}\)

Now, substitute the values into the angular frequency formula:

\(\omega = \sqrt{\frac{150}{\frac{1}{6}}} = \sqrt{150 \times 6} = \sqrt{900}\)

\(\omega = 30\,\text{rad/s}\)

However, it seems there is an error in the calculated value. Let's re-evaluate:

\(\omega = \sqrt{\frac{150}{\frac{1}{6}}} = \sqrt{150 \times 6} = \sqrt{900} = 30\,\text{rad/s}\)