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प्रश्न
The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
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उत्तर
As the block of mass M is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.
Now, springs k2 and k3 are in connected as a series combination.
Let k4 be the equivalent spring constant.
\[\therefore \frac{1}{k_4} = \frac{1}{k_2} + \frac{1}{k_3} = \frac{k_2 + k_3}{k_2 k_3}\]
\[ k_4 = \frac{k_2 k_3}{k_2 + k_3}\]
k4 and k1 form a parallel combination of springs. Hence, equivalent spring constant k = k1 + k4.
\[= \frac{k_2 k_3}{k_2 + k_3} + k_1 \]
\[ = \frac{k_2 k_3 + k_1 k_2 + k_1 k_3}{k_2 + k_3}\]
\[ \therefore \text { Time peiod }, T = 2\pi\sqrt{\frac{M}{k}}\]
\[ = 2\pi\sqrt{\frac{M \left( k_2 + k_3 \right)}{k_2 k_3 + k_1 k_2 + k_1 k_3}}\]
(b) Frequency \[\left( v \right)\] is given by,
\[v = \frac{1}{T}\]
\[= \frac{1}{2\pi}\sqrt{\frac{k_2 k_3 + k_1 k_2 + k_1 k_3}{M\left( k_2 + k_3 \right)}}\]
(c) Amplitude ( x ) is given by,
\[x = \frac{F}{k} = \frac{F\left( k_2 + k_3 \right)}{k_1 k_2 + k_2 k_3 + k_1 k_3}\]
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