Question

the potential energy U is given by equation U=(2.5x²+20)joule. the mass of the particle is 0.2kg ,then​

Answer 1

Answer:

Explanation:

Given:

U = 2.5x²+ 20 joule

m = 0.2 kg

Calculation:

Step 1:

As we know that in mean position the potential energy is equal zero.

At SHM, x=2

U=2.5×2²+20

U= 10+20

U=30 J

Step 2:

Therefore, we see than mean position is not equal to zero. Hence option a is incorrect.

Now, check for the time period of oscillation is 2.5 second.

As we know that Force ,F=$\frac{-dU}{dx}$

F = $\frac{-d}{dx}$(2.5x²+ 20)

F= -5x

we can write it as; k = 5 N/m

Step 3:

Angular frequency,ω=$\sqrt{\frac{k}{m} }$

ω=$\sqrt{25}$

ω= 5 rad/s

T =√2π/ω

T=√2×3.14/5

T= 1.25 sec

Result:

Hence angular frequency of oscillation is 5 rad$s^{-1}$

Option (c) is correct.

Answer 2

Potential energy, $\mathrm{U}=\left(2.5 \mathrm{x}^{2}+20\right)$

Mass, $\mathrm{m}=0.2 \mathrm{~kg}$

Time,$t=?$

Mean position$=?$

We know that

Potential energy,$U(x)=\left(2.5 x^{2}+20\right) \mathrm{J}$

The force acting on particle,

$\Rightarrow \mathrm{F}=-\mathrm{du} / \mathrm{dx}$

$\Rightarrow F=-\left[d / d x\left(2.5 x^{2}+20\right)\right]$

$\Rightarrow \mathrm{F}=-2.5 \times 2 \mathrm{x}$

$\Rightarrow \mathrm{F}=-5 \mathrm{x}$.............(i)

Now, Acceleration

So $\mathrm{F}=\mathrm{ma}$

$\Rightarrow-5 \mathrm{x}=0.2 \times \mathrm{a} \quad$.........from (i)

$\Rightarrow a=-5 / 0.2 x$

$\Rightarrow \quad \mathrm{a}=-25 \mathrm{x}$............(ii)

Then we can find the time from angular frequency,

$\Rightarrow a=-\omega^{2} x$

$\Rightarrow-25 x=-\omega^{2} x$.................from (ii)

$So, w=5$.........(iii)

We know that $\omega=2 \pi / \mathrm{t}$

$\Rightarrow \mathrm{T}=2 \pi / 5 \quad$......... from (iii)

$\Rightarrow \mathrm{T}=1.25 \mathrm {sec}$

As the mean position ,force is acting on particle is zero

$F=-5 x$

$So, \omega=5$

We know that $\omega=2 \pi / t$

$\Rightarrow T=2 \pi / 5 \quad$......... from (iii)

$\Rightarrow T=1.25 \mathrm{sec} \quad$ (iii)

As the mean position , force is acting on particle is zero

$F=-5 x$

$\\ 0=5 x \\ x=0$

So, the mean position of particle executing SHM is $x=0 \mathrm{~m}$

Angular frequency of oscillation is $5 rads^{-1}$