Question

expression for thr period of oscillation of a spring and explain the terms​

Answer 1

Answer:

Given:

A bob is attached to a spring and allowed to oscillate.

To find:

Equation for the time period and Explanation of the terms.

Calculation:

Equation of motion for the spring attached to a mass "m"

F = - kx

=> ma = - kx

=> a = - (k/m) x

So time period

$T = 2\pi \sqrt{ \frac{m}{k} }$

Explanation:

The main terms in the Equation are as follows :

1. T => time period

2. m => mass of the object attached to the spring

3. k => spring constant . It's an index of oscillations of the spring.

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

Expression of time period of a spring :

$\LARGE \implies {\boxed{\sf{F \: \propto \: - x}}}$

Where, F is force and x is the displacement of the spring,

$\Large \rightarrow {\sf{F \: = \: -kx}}$

Where,

k is spring constant

$\Large \rightarrow {\sf{ma \: = \: -kx \: \: \: \: \: [\because \: F \: = \: ma]}}$

So, we can write,

$\Large \rightarrow {\sf{a \: = \: \frac{-kx}{m}}}$

$\Large \rightarrow {\sf{ \cancel{-} \omega ^2 \cancel{x} \: = \: \frac{\cancel{-}k \cancel{x}}{m} \: \: \: \: [\because \: a \: = \: - \omega ^2 x]}}$

$\Large \rightarrow {\sf{\omega \: = \: \sqrt{\frac{k}{m}}}}$

$\Large \rightarrow {\sf{\frac{2 \pi}{T} \: = \: \sqrt{\frac{k}{m}} \: \: \: \: [\because \: \omega \: = \: \frac{2 \pi}{T}]}}$

$\LARGE \implies {\boxed{\boxed{\sf{T \: = \: 2 \pi \sqrt{\frac{m}{k}}}}}}$

Where,

T is time period

k is spring constant

m is mass

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