Figure (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 (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) is released, what is the period of oscillation in each case?

Asked by Abhisek | 1 year ago |  82

Solution :-

(a) The maximum extension of the spring in fig (a) is x.  The F = kx

Therefore, x =$$\dfrac{F}{k}$$

The force on each mass acts as the reaction force acting on the other mass. The two mass behaves as if it is fixed with respect to the other.

Therefore, x = $$\dfrac{F}{k}$$

here k is the spring constant

(b)  In figure (a) the restoring force on the mass is F = -kx, here x is the extension of the spring.

For the mass (m) of the block, force is written as

F = ma = m ($$\dfrac{d^2x}{dt^2}$$)

Therefore, m ($$\dfrac{d^2x}{dt^2}$$) = -kx

($$\dfrac{d^2x}{dt^2}$$) = ($$\dfrac{-k}{m}$$)x = – ω2x

Here, angular frequency of oscillation, ω = $$\sqrt{\dfrac{k}{m}}$$

Time period of oscillation, T =$$\dfrac{2π}{ω}$$

= 2π ($$\sqrt{\dfrac{m}{k}}$$)​

In Figure (b), the centre of the system is O and there are two springs.

Each spring is of length $$\dfrac{I}{k}$$ and it is attached to two masses

Therefore F = – 2kx

here x is the extension of the spring.

F = ma = m ($$\dfrac{d^2x}{dt^2}$$)

m ($$\dfrac{d^2x}{dt^2}$$) = -2kx

($$\dfrac{d^2x}{dt^2}$$) = ($$\dfrac{-2k}{m}$$)x= -ωx

​ω = ($$\sqrt{\dfrac{2k}{m}}$$)

T = 2π ($$\sqrt{\dfrac{m}{2k}}$$)

Answered by Pragya Singh | 1 year ago

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