Question 52: Oscillations

Question

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be

Accepted Answer

Given:
Natural length of the spring: 2 meters
Initial compression of the spring ($ x_i $): 1 meter
Final compression of the spring when the block is at distance $ x $ ($ x_f $): $ (2 - x) $ meters
Solution:
To find the block's speed, we'll use energy conservation. Initially, the spring's potential energy is fully converted to kinetic energy and spring potential energy at the new position $ x $.
Conservation of Energy:
Initial kinetic energy ($ K_i $) is 0 since the block starts from rest.
Initial potential energy ($ U_i $) due to spring compression is $\frac{1}{2} K x_i^2$.
At a distance $ x $, the kinetic energy ($ K_f $) is $\frac{1}{2} mv^2$ and the potential energy ($ U_f $) is $\frac{1}{2} K x_f^2$.
$    K_i + U_i = K_f + U_f    $
$    0 + \frac{1}{2} K x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} K x_f^2    $
Rearrange to find $ v^2 $:
$    \frac{1}{2} m v^2 = \frac{1}{2} K (x_i^2 - x_f^2)    $
Substitute known values:
$    \frac{1}{2} 	imes 2 	imes v^2 = \frac{1}{2} 	imes 200 	imes \left(1^2 - (2-x)^2 ight)    $
Simplify and solve for $ v^2 $:
$    v^2 = 100 \left[ 1 - (2-x)^2 ight]    $
Finally, take the square root to find $ v $:
$    v = 10 \left[ 1 - (2-x)^2 ight]^{1/2}    $
Therefore, the speed of the block at distance $ x $ from the wall is given by the expression above.

Answer

$10\left[1-(2-x)^2\right]^{\frac{1}{2}} \ m/s$

Answer

$10\left[1-(2-x)^2\right]^{\frac{3}{2}} \ m/s$

Answer

$10\left[1-(2-x)^2\right] \ m/s$

Answer

$10\left[1-(2-x)^2\right]^2 \ m/s$

Oscillations

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