Question

Q. 2.
(iii) A body of mass 2 kg performs linear SHM. The restoring force acting on it is
3 N when it is 0.06 m from the mean position.
The differential equation of its motion is
d²x
d²x
+ 100x = 0
(b)
(a)
d12
dtr + 25x=0
d²x
(C) 50
+x=0
(d) 2 dr
+ 3x = 0.
the differential equations of its motion ​

Answer 1

Answer:

2d²x+3x=0 it is the answer of this question

____

dt²

Answer 2

The differential equation of the motion of the body in linear SHM is $\frac{d^2x}{dt^2}+25x=0$

Explanation:

• For a body of mass 'm' , displaced by 'Δx' from its mean position and having a restoring force 'F' acting on it is, performing a linear SHM (simple harmonic motion) then the differential equation of its motion is given by,                                                                                               [tex]\frac{d^2x}{dt^2} +\omega^2x=0\\ ->F=-K x\\ ->\omega^2=\frac{K}{m} [/tex]                  
• here we have $m=2kg\ \ \ \ \ \ \ F=3N\ \ \ \ \ \ \ \ \ \Delta x=0.06m$                                      hence, equating these for expression for acceleration we get,                           [tex]->F=-K x\\ ->ma=-K x\\ ->a=-\frac{K}{m}x\\ ->\frac{d^2x}{dt^2}+\frac{K}{m}x= 0[/tex]          -----(a)
• now for the given displacement we have,                                                    [tex]F=K\Delta x\\ K=\frac{\Delta x}{F}\\ K= \frac{3}{0.06}\ \ \ \ \ ->K=50 [/tex]  
• putting these in (a) we get,  $\frac{d^2x}{dt^2}+\frac{50}{2}x= 0 \ \ \ \ ->\frac{d^2x}{dt^2}+25x= 0$  (ans)