First slide

Simple hormonic motion

Question

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]

Difficult

A

5 m
B

3 m
C

7 m
D

8 m

Solution

$U = 2 - 20 x + 5 x^{2}$
$F = - \frac{dU}{dx} = 20 - 10 x$
At equilibrium position; F = 0
20 - 10x = 0
$\Rightarrow$ x =2
It will oscillate about x = 2 m.
Now m $\overset{. .}{m x = F = - 10 (X - 2)} \Rightarrow \overset{. .}{x} + 10 (x - 2) = 0 . T h e a b o v e e q u a t i o n r e p r e s e n t s a n S H M w h o s e e q u a t i o n i s g i v e n b y x - 2 = A \cos (ω t + φ) A t t = 0, x = 3 m a n d a t t = 0, v = \frac{d x}{d t} = 0 S o l v i n g A = 1 m a n d φ = 0 . T h e r e f o r e x - 2 = 1 \cos ω t . S i n c e m a x i m u m v a l u e o f \cos ω t i s + 1, w e g e t x_{m a x} = (2 + 1) m = 3 m .$
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