Question

One end of a light spring of force constant k is fixed to wall and the other end is tied to a block placed on a smooth

Answer 1

"Amplitude of the motion, $A = 2X_0$

We know that the time needed to cover from compressed position to mean position (normal) $= \frac {T}{4}$

Where, T is the time period of oscillation.

For the displacement from the mean position to $X_0$, the time taken can be obtained from:

$Y = A sin \omega t$; where Y = displacement from the mean position and A is the amplitude.

Therefore : $X_0 = 2 X_0 sin \omega t$; but $\omega = \frac {2\pi}{T}$

$1/2= sin \frac {2\pi}{T} t$      

$\frac {2\pi}{T} t = sin-1 (0.5)$

$\frac {2\pi}{T} t = \frac {\pi}{6}$

$t = \frac {T}{12}$

Therefore the time taken to hit the wall be  $= \frac {T}{4} +\frac {T}{12}$

$= \frac {T}{3}$

If a mass M is suspended from a spring of force constant k, then the time period will be:

$T = 2 \pi \sqrt {\frac{M}{K}}$

Time taken to hit the wall $=\frac {T}{3}= 2 \pi \sqrt {\frac{M}{K}}$"