Question

Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring constant k2. Each spring is attached on the other side to a wall, and the block slides without friction on a horizontal surface. When the block is sitting at x = 0, both springs are relaxed. Write Newton’s second law, F = ma, as a differential equation for an arbitrary position x of the block. What is the period of oscillation of this system?

Answer 1

Answer:

$T= 2\pi\times sqrt(m/(k1+k2))$

Explanation:

When the block is displaced by x units

F= spring force

two springs are connected parallel

$F =-k_1x - k_2x$

Writing Newtons second law, F = ma

$-k_1x - k_2x =ma$

$-k_1x - k_2x = mx''$

a= x" ( differentiating x w.r.t time twice)

$x''+(k_1/m + k_2/m) x=0$

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

since,  Period $T=2\pi\sqrt{\frac{m}{K_{eq.}} }$

therefore,

$T= 2\pi\times sqrt(m/(k1+k2))$