Question

Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.30(b) 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 Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer 1

(i) we know,
|F| = Kx

Let x' is the extension in the spring then , F = Kx'
x' = F/K

In both cases maximum extension in the spring is (F/K).

(ii) for case (a)
F = - kx
Where x is extension
But a/c to Newton's 2nd law .
F = ma
ma = -Kx
a = -(K/m)x
But we know, for SHM
a = -w²x
Where w is angular velocity
w = √(k/m)
2π/T = √(K/m)
T = 2π√(m/K)
Hence, period of 1st case is 2π√(m/K)

For cose (b)
The system is divided into two similar systems with spring divided in two equL halves . K' = 2K
So, F = -K'x = -2Kx
From Newton's 2nd law,
F = ma
So, - 2Kx = ma
a = - (2K/m)x
Compare with a = -w²x
w = √(2K/m)
2π/T = √(2K/m)
T = 2π√(m/2K)

Answer 2

Answer:

Hope these will help all of u guys