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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-14 Oscillations



Q.13:- 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:-
(a) The maximum extension of the spring in both cases will = Flk, where k is the spring constant of the springs used.
(b) In Fig.14.30(a) if x is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass iss F = –kx, i.e., F ∝ x
As, this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM.
Spring factor = spring constant = k
inertia factor = mass of the given mass = m
As time period,
In Fig.14.30(b), we have a two body system of spring constant k and reduced mass, µ = m × m / m + m = m/2.
Inertia factor = m/2
Spring factor = k