Question

Figure below shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. If force F applied at the free end stretches the spring. Figure shows the same spring with both ends free and attached to a mass m at the either end. Each end of the spring in figure is stretched by the same force F.a. What is the maximum extension of the spring in the two cases?b. If the mass in figure(a) and two masses in figure(b) are released, what is the period of oscillation in each case?

Answer 1

(a)
we know, restoring force in a spring is given as
$F_s$ = Kx
Where K is the spring constant of the spring and x is the extension in the spring.
So, F = Kx
i.e. x = F/K

hence, maximum extension of string in case (a) is F/K.

now in case (b) restoring force on each block as shown in figure. here you can see restoring force equals force applied on it.
so, Fs = Kx = F
x = F/K
hence, maximum extension of string is F/K in case(b)

(b) F = -Kx [ here negative sign shows force is in opposite direction to displacement ]
we know, motion of spring system is a simple harmonic motion,
so, $a=-\omega^2x$ ......(1)

from Newton's 2nd law,
F = ma = -Kx
a = -(K/m)x .........(2)

compare equations (1) and (2),
we get, $\omega^2=\frac{K}{m}$
so, $T=2\pi\sqrt{\frac{m}{K}}$

Now in Case (b) , both blocks will move to and fro from each other and start oscillating simple harmonically and the mean position will be different for both blocks on either side of centre of spring as shown in the figure

Now the spring can be divided in two parts and we can $x_0$ is the natural length of both part of spring on either side, now considering any one Block we can say time period of oscillation is $T=2\pi\sqrt{\frac{m}{K'}}$
Where m is the mass of the block and K’ is the spring constant for that part

we know, $\textbf{When a spring is cut to half }\\\textbf{value of spring constant doubles}$

So if K is the original spring constant, Spring constant of each part is
K’ = 2K,

hence, $T=2\pi\sqrt{\frac{m}{2K}}$