Question

Combine newton's 2nd law and hooke's law for a spring to find the acceleration of the block a(t as a function of time.

Answer 1

2nd law: F = ma 
hook's law: F = -kx

$ma = -kx$
$a = \frac{d^2x}{dt^2} = - \frac{k}{m} x$

try solution: $x = e^{ \lambda t} , \frac{dx}{dt} = \lambda e^{ \lambda t}, \frac{d^2x}{dt^2} = \lambda^2 e^{ \lambda t}$

$\lambda^2e^{ \lambda t}= - \frac{k}{m}e^{ \lambda t}$

divide by: $e^{ \lambda t}$

$\lambda^2 = - \frac{k}{m}$

solution:
$a =-\frac{k}{m}e^{i \sqrt{ \frac{k}{m} }t }$