Question

The potential energy of a particle of mass 100 g moving along was is given by U=5x(x-4) Where X in meter
mere The period of oscation is​

Answer 1

Answer:

• The Time period (T) of the oscillation is π / 5 sec

Given:

1. Mass of particle = 100 g = 0.1 Kg
2. Given Equation ; U = 5 x(x - 4)

Explanation:

$\rule{300}{1.5}$

From the given equation,

$\displaystyle\Rightarrow\sf U=5 x\;(x-4)\\\\\\\Rightarrow\sf U=5\;x^{\;2} - 20\;x$

Differentiating the above equation with respect to displacement to get Force.

Therefore,

$\displaystyle\Rightarrow\sf \dfrac{d\;U}{d\;x}=\dfrac{d(5\;x^{\;2}-20\;x)}{d\;x}\\\\\\\Rightarrow\sf F=\dfrac{d(5\;x^{\;2}-20\;x)}{d\;x}\;\;\;\left(F=\dfrac{d\;U}{d\;x}\right)\\\\\\\Rightarrow\sf F=10\;x-20\\\\\\\Rightarrow\sf F=10\;(x-2)$

• Let (x - 2) = X

$\displaystyle\Rightarrow\sf F=10\;(x-2)\\\\\\\Rightarrow\sf F=10\;X\;\;\;\bigg((x-2)=X\bigg)$

Comparing with the Equation,

$\sf F=K\;x$

∴ We get the Force Constant (K) as 10.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Time Period Formula we know,

$\bullet\;\underline{\boxed{\sf T=2\;\pi\;\sqrt{\dfrac{m}{K}}}}$

Here,

• T Denotes Time Period.
• m Denotes Mass.
• K Denotes Force Constant.

Substituting the values,

$\displaystyle\Rightarrow\sf T=2\;\pi\;\sqrt{\dfrac{0.1}{10}}\\\\\\\Rightarrow\sf T=2\;\pi\;\sqrt{\dfrac{1}{10\times 10}}\\\\\\\Rightarrow\sf T=2\;\pi\times\dfrac{1}{10}\\\\\\\Rightarrow\sf T=\dfrac{\pi}{5}\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf T=\dfrac{\pi}{5}\;Sec}}}}$

∴ The Time period (T) of the oscillation is π / 5 sec.

$\rule{300}{1.5}$