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

Concept: When two masses are connected by a spring and allowed to oscillate on a frictionless surface, the system executes simple harmonic motion about the centre of mass. The effective mass \(\mu\) of the system is the reduced mass: \[ \mu=\frac{m_1 m_2}{m_1+m_2} \] The angular frequency of oscillation is: \[ \omega=\sqrt{\frac{k}{\mu}} \] 
Step 1: Identify given data \[ m_1=1\,\text{kg},\quad m_2=0.2\,\text{kg},\quad k=150\,\text{N/m} \] 
Step 2: Calculate reduced mass \[ \mu=\frac{(1)(0.2)}{1+0.2} =\frac{0.2}{1.2} =\frac{1}{6}\,\text{kg} \] 
Step 3: Calculate angular frequency \[ \omega=\sqrt{\frac{k}{\mu}} =\sqrt{\frac{150}{1/6}} =\sqrt{900} =30 \] But note that the angular frequency of relative oscillation is \(30\), while the angular frequency of each mass about the centre of mass is: \[ \omega=\sqrt{\frac{k}{m_1+m_2}} =\sqrt{\frac{150}{1.2}} =\sqrt{125} \approx 11.18 \] The standard result used in such problems (oscillation of separation between masses): \[ \omega=\sqrt{\frac{k(m_1+m_2)}{m_1 m_2}} =\sqrt{\frac{150\times1.2}{0.2}} =\sqrt{900} =30 \]