Question

If a mass m is suspended with a spring then we get x extension. Now we attech an another mass (M)

with the previous one and again oscillate, was what will be the time period of the osullations of the

system?​

Answer 1

Answer:

mg=kx

∴   k=  mg/x

​  

 

T=2π  

k

M+m

​  

 

​  

=2π  

mg

(M+m)x

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Explanation:

i think soo this will be the answer

Answer 2

Answer:

Period of oscillation $\mathbf{T}=2 \pi \sqrt{(M+m) x} / \mathrm{mg})$

Explanation:

Given: Two masses $M and m$ are suspended on a massless spring. Distance $x$ is the stretch by $m$.

Find: Period of oscillation of the combined mass.

Solution:

$\mathrm{mg}=\mathrm{kx}$

where $\mathrm{m}$ is the mass, $\mathrm{g}$ is gravity, $\mathrm{x}$ is the stretch and $\mathrm{k}$ is the spring constant.

Therefore $\mathbf{k}=\mathbf{m g} / \mathbf{x}$

Period of oscillation is derived from the formula:

$\begin{aligned}&T=2 \pi V(M+m / k) \\&\mathbf{T}=\mathbf{2 \pi} \sqrt{(M+m) x} / \mathbf{m g})\end{aligned}$