physics ncert solutions class 11th Physics Class 11th Oscillations

14.13 Figure 14.26 (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 14.26 (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. 14.26(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 ?

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Vishal Baghel | Contributor-Level 10

5 months ago

(a) For figure (a) : When a force F is applied to the free end of the spring, an extension l is produced. For the maximum extension, it can be written as:

F – kl, where k is the spring constant.

For maximum =extension of the spring, l = $\frac{F}{k}$

For figure (b): The displacement (x) produced in this case is x = $\frac{1}{2}$

Net force F = +2kx = 2k $\frac{1}{2}$ . So l = $\frac{F}{k}$

(b) For figure (a) : For mass (m) of the block, force is written as : F = ma = m $\frac{d^{2} x}{d t^{2}}$ ,

where x is the displacement of the block in time t, then

m $\frac{d^{2} x}{d t^{2}} = - k x$ , it is negative because the direction of the elastic force is opposite to the direction of displacement.

$\frac{d^{2} x}{d t^{2}} = - (\frac{k}{m}) x$ &nbs

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(a) For figure (a) : When a force F is applied to the free end of the spring, an extension l is produced. For the maximum extension, it can be written as:

F – kl, where k is the spring constant.

For maximum =extension of the spring, l = $\frac{F}{k}$

For figure (b): The displacement (x) produced in this case is x = $\frac{1}{2}$

Net force F = +2kx = 2k $\frac{1}{2}$ . So l = $\frac{F}{k}$

(b) For figure (a) : For mass (m) of the block, force is written as : F = ma = m $\frac{d^{2} x}{d t^{2}}$ ,

where x is the displacement of the block in time t, then

m $\frac{d^{2} x}{d t^{2}} = - k x$ , it is negative because the direction of the elastic force is opposite to the direction of displacement.

$\frac{d^{2} x}{d t^{2}} = - (\frac{k}{m}) x$ = ${- ω}^{2}$ xwhere, $ω^{2}$ = $\frac{k}{m}$ , $ω i s a n g u l a r f r e q u e n c y o f t h e o s c i l l a t i o n .$

Time period of the oscillation, T = $\frac{2 π}{ω}$ = $\frac{2 π}{\sqrt \frac{k}{m}}$ = 2 $π \sqrt{\frac{m}{k}}$

For figure (b) :

F = m $\frac{d^{2} x}{d t^{2}}$ = -2kx. It is negative because the direction of elastic force is opposite to the direction of displacement.

$\frac{d^{2} x}{d t^{2}} = - [\frac{2 k x}{m}] x$ = - $ω^{2}$ x, where angular frequency $ω$ = $\sqrt{\frac{2 k}{m}}$

Time period, T = $\frac{2 π}{ω} = 2 π \sqrt{\frac{m}{2 k}}$

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