Question

Fig. (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. Fig. (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 (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 free, what is the period of oscillation in each case?

Answer 1

(a) If l is the extension produced in the spring then elastic force developed in the spring is balanced by applied force F.

We have F = kl,

or, l = F/k   ....(i)

Again F = ma = md²y/dt²

Hence, md²y/dt² = -ky  [displacement of mass, y = l]

(Negative sign has been introduced as the direction of elastic force is opposite to the direction of displacement y).

or, d²y/at² = -(K/m)y = -w²y    ....(iii)

where ω² = k/m

or, ω = √{k/m}    ....(iv)

Equation (iii) shows that motion of mass 'm' is simple harmonic whose time period is given by:

T = 2π/ω = 2π/√{k/m} = 2π√{m/k}   ...(v)

Further from equation (i) extension produced in the spring,

l = F/k

(b) In this case displacement of each mass,

y = 1/2

or displacement, y = {F/k}/{2} = F/2k

or, F = 2ky   ...(vi)

Also, as F = m.d²y/dt²   ....(vii)

Hence, from (vi) and (vii),

m.d²y/dt² = -2ky

(Again, negative sign is introduced as the direction of elastic force is opposite to direction of displacement y).

or, d²y/dt² = -(2k/m)y

= ω²y   ....(viii)

where ω² = 2k/m

or, ω = √{2k/m}

Thus, we again find from equation (viii) that here also masses execute simple harmonic motion, whose time period is given by

T = 2π/ω = 2π/√{2k/m}

= 2π m/2k