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3 years ago

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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 con

stant 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?

1 answer:

Andrew [12]3 years ago

3 0

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,

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

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