Solution: Block Suspended by Springs and Pulley
Let the block move down by a distance \( y \). Since the block hangs from the pulley, the pulley also moves down by \( y \).
For the pulley to move down by \( y \), the loop of the cord passing under it must lengthen by \( 2y \). This length comes from the extension of the two springs.
$$ x_1 + x_2 = 2y $$where \( x_1 \) and \( x_2 \) are the extensions of springs \( k_1 \) and \( k_2 \).
Since the cord is light and the pulley is ideal, the tension \( T \) is uniform throughout the cord.
Forces on the springs:
$$ k_1 x_1 = T \implies x_1 = \frac{T}{k_1} $$ $$ k_2 x_2 = T \implies x_2 = \frac{T}{k_2} $$Substitute these into the constraint equation:
$$ \frac{T}{k_1} + \frac{T}{k_2} = 2y \implies T \left( \frac{k_1 + k_2}{k_1 k_2} \right) = 2y $$ $$ T = \left( \frac{2 k_1 k_2}{k_1 + k_2} \right) y $$The total upward restoring force on the block (supported by the two segments of the cord) is \( F_{restoring} = 2T \).
$$ F_{restoring} = 2 \left[ \left( \frac{2 k_1 k_2}{k_1 + k_2} \right) y \right] = \left( \frac{4 k_1 k_2}{k_1 + k_2} \right) y $$This follows the form \( F = K_{eff} y \), where:
$$ K_{eff} = \frac{4 k_1 k_2}{k_1 + k_2} $$The period is \( T = 2\pi \sqrt{\frac{m}{K_{eff}}} \).