A particle is moving on x -axis has potential energy U=(2 - 20 x + 5 x2) joules along x -axis. The particle is released at x=-3. The maximum value of 'x' will be [ x is in meters and U is in joules]:

Question

A particle is moving on x -axis has potential energy U=(2 - 20 x + 5 x2) joules along x -axis. The particle is released at x=-3. The maximum value of 'x' will be [ x is in meters and U is in joules]: (A) 5m (B) 3m (C) 7m (D) 8m

Tardigrade Admin · Accepted Answer

U=2-20x+5x2 F=-(d u/d x)=20-10x[.F propto -x⇒ SHM]. At equilibrium F=0 20-10x=0 x=2 [mean position] since particle is released at x=-3, therefore amplitude of particle is 5 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-krphhdtajdp6yp3g.jpg" />

Q. A particle is moving on $x$ -axis has potential energy $U = (2 - 20 x + 5 x^{2})$ joules along $x$ -axis. The particle is released at $x = - 3.$ The maximum value of $^{'} x^{'}$ will be [ $x$ is in meters and $U$ is in joules]:

$5 m$

$3 m$

$7 m$

$8 m$

$U = 2 - 20 x + 5 x^{2}$
$F = - \frac{d u}{d x} = 20 - 10 x [F \propto - x \Rightarrow S H M]$
At equilibrium $F = 0$
$20 - 10 x = 0$
$x = 2$ [mean position]
since particle is released at $x = - 3,$ therefore
amplitude of particle is $5$

Q. A particle is moving on x -axis has potential energy U=(2−20x+5x2) joules along x -axis. The particle is released at x=−3. The maximum value of ′x′ will be [ x is in meters and U is in joules]:

5m

3m

7m

8m