The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended, the period of oscillation will now be

Question

What is the fundamental frequency of an open organ pipe with length $ L $ and speed of sound $ v $?

Answer 1

To determine the new period of oscillation when another mass $M$ is added to the spring system, we begin by recalling the formula for the period of a spring-mass system. The period $T$ of oscillation for a mass $M$ suspended from a spring is given by:

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

where $k$ is the spring constant.

When another mass $M$ is added, the total mass becomes $2M$. The new period of oscillation $T_{\text{new}}$ is:

T_{\text{new}} = 2\pi \sqrt{\frac{2M}{k}}

We can express the new period in terms of the original period $T$ by factoring out 2 from the expression:

T_{\text{new}} = 2\pi \sqrt{\frac{2M}{k}} = 2\pi \sqrt{2 \times \frac{M}{k}} = 2\pi \sqrt{2} \sqrt{\frac{M}{k}} = \sqrt{2} \times (2\pi \sqrt{\frac{M}{k}}) = \sqrt{2} \times T

Therefore, the period of oscillation when the mass is doubled is $ \sqrt{2} T $.

This means the correct answer is $ \sqrt{2} T $, making the correct option: $ \sqrt{2} T $.

Answer 2