Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by 0.065 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.048 m relative to its unstrained length

Answer 1

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = $= \omega^2 \times m$

k = (11.3)² m

k = 127.7 m

where;

$x_1$ = 0.065 m

$x_2$  = 0.048 m

According to the conservation of energies;

$E_1=E_2$

∴

$\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2$

$kx_1^2 = mv_2^2 + kx_2^2$

$(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2$

$0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325 - 0.2942208 = v_2^2 \\ \\ v_2^2 = 0.2453117 \\ \\ v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}$