Question

a body of mass 2kg attached to a spring is displaced through 4cm from it's equilibrium position and then released .Calculate the frequency of vibration if spring having 200 N/m
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Answer 1

Answer:

Given:

Mass = 2kg

Displacement = 4 cm

Spring constant = 200 N/m

To find:

Frequency of Vibration

Concept:

When the mass was displaced by 4cm , it acts as the amplitude for Oscillation of the spring.

So amplitude = 4cm = 0.04 m

Frequency is the total number of Oscillations made in 1 sec. In other words , it's the reciprocal of time period.

Calculation:

Frequency be denoted as f

$f = \dfrac{1}{2\pi} \sqrt{ \dfrac{k}{m} }$

$= > f = \dfrac{1}{2\pi} \sqrt{ \dfrac{200}{2} }$

$= > f = \dfrac{1}{2\pi} \sqrt{ 100}$

$= > f = \dfrac{1}{2\pi} \times 10$

$= > f = \dfrac{5}{\pi}$

$= > f = 1.59 \: hz$

So final answer :

$\boxed{ \huge{ \green{ \sf{ f = 1.59 \: hz}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Mass of Body (m) = 2 kg
• Springs Constant (k) = 200 N/m
• Displacement (x) = 4 cm

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To Find :

• Frequency of vibration produced in spring.

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Solution :

We have formula for Time Period of Simple Pendulum or of spring :

$\large{\boxed{\sf{T \: = \: 2 \pi \sqrt{\dfrac{m}{k}}}}} \\ \\ \implies {\sf{T \: = \: 2 \pi \sqrt{\dfrac{2}{200}}}} \\ \\ \implies {\sf{T \: = \: 2 \pi \sqrt{\dfrac{1}{100}}}} \\ \\ \implies {\sf{T \: = \: 2 \pi 0.01}} \\ \\ \implies {\sf{T \: = \: 0.02 \pi}}$

And we know that frequency is reciprocal of Time Period.

$\large{\boxed{\sf{f \: = \: \dfrac{1}{T}}}} \\ \\ \implies {\sf{f \: = \: \dfrac{1}{0.02 \pi}}} \\ \\ \implies {\sf{f \: = \: \dfrac{1}{0.02 \: \times \: 3.14}}} \\ \\ \implies {\sf{f \: = \: \dfrac{1}{0.0628}}} \\ \\ \implies {\sf{f \: = \: 15.9}} \\ \\ \underline{\sf{\therefore \: Frequency \: is \: 15.9 \: Hz}}$