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Question

Physics Question on simple harmonic motion

Two blocks each of mass mm is connected to the spring of spring constant kk as shown In the figure. If the blocks are displaced slightly In opposite directions and released, they will execute simple harmonic motion. The time period of oscillation is

A

2πmk2\pi\sqrt{\frac{m}{k}}

B

2πm2k2\pi\sqrt{\frac{m}{2k}}

C

2πm4k2\pi\sqrt{\frac{m}{4k}}

D

2π2mk2\pi\sqrt{\frac{2m}{k}}

Answer

2πm2k2\pi\sqrt{\frac{m}{2k}}

Explanation

Solution

Reduced mass of the system μ=(m)(m)m+m=m2\mu = \frac{\left(m\right)\left(m\right)}{m+m} = \frac{m}{2} \therefore Time period, T=2πμkT= 2\pi\sqrt{\frac{\mu}{k}} =2πm2k= 2\pi\sqrt{\frac{m}{2k}}