Question

Prove the correctness of this equation T=2π√m/k

Answer 1

 Best Answer

you need to define your terms, not make me guess at them. However, I do recognize the equation as applying to a pendulum. The full equation is: 

Pendulum period in seconds 
T ≈ 2π√(L/g) 
L is length of pendulum in meters 
g is gravitational acceleration = 9.8 m/s² 

L/g = m / (m/s²) = s² 
√(L/g) = s 
2π is dimensionless 
therefore T is in seconds

Answer 2

Explanation:

To prove: $T=2\pi \sqrt{m/k}$

T is Time period of spring in SHM.

Step by step explanation:

Applying newton's law:

$F=-kx=ma$

Here,

F= Spring Force

k= Spring constant

In SHM:   $a=-\omega ^{2}x$

$a\propto -x\\\\a=-\omega ^{2}x\rightarrow equation\ 1\\-kx=ma\rightarrow equation\ 2\\k=spring\ constant\\\\a=-kx/m=-\omega ^{2}x\\\omega ^{2}=k/m\\Since\ T=2\pi /\omega \\\omega =\sqrt{k/m}\\T=2\pi /\sqrt{k/m}=2\pi\sqrt{m/k}$

Dimension of time period is M⁰L⁰T¹ which is thus obtained.

Therefore, the formula for time period is correct which is : $T=2\pi \sqrt{m/k}$

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