In the case of simple harmonic motion of spring block system, why time period of the simple harmonic motion of the block is independent of acceleration of the system (spring-block system)?
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A spring applies a force $F = kx$, where $k$ is the spring constant and $x$ is the distance from the rest position. If you have a mass $m$ on the spring, it will oscillate around $x = 0$ with period $t = 2\pi\sqrt{m\over{k}}$. If you then apply an acceleration $a$, the rest position of the spring will shift by $d_a = {a\over{k}}$. If you keep this acceleration constant, and displace the spring away from its new rest position $d_a$, then the net spring force will be $F_n = k(x - d_a)$. This is the same equation as you had before for the spring, just shifted by $d_a$. So, you will again get oscillation with period $t = 2\pi\sqrt{m\over{k}}$, except centered on $x = d_a$.
answered Oct 13, 2015 at 1:52
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$\begingroup$ Why this shift causes no change in time period? Please explain. Thanks in advance $\endgroup$ Commented Oct 13, 2015 at 3:24
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$\begingroup$ The oscillation period is a function of two things: the restoring force as a function of displacement from the spring's rest position, and the mass. Neither changes if you add an acceleration; you just shift the spring's rest position. $\endgroup$ Commented Oct 13, 2015 at 3:28