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 0.060 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 12 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length?

Answer 1

Answer:

Velocity = 0.4762 m/s

Explanation:

Given the details for the simple harmonic motion from the question as:

Angular frequency, ω = 12 rad/s

Amplitude, A = 0.060 m

Displacement, y = 0.045 m

The initial Energy =  U  = (1/2) kA²    

where A is the amplitude and k is the spring constant.

The final energy is potential and kinetic energy

   K + U =   (1/2) mv²   + (1/2) kx²  

where  x  is the displacement

m is the mass of the object

v is the speed of the object

Since energy is conservative. So, the final and initial energies are equal  as:

   (1/2) k A²   = (1/2) m v²   + (1/2) kx²  

Using,   ω² = k/m, we get:  

Velocity:

$v=\omega\times \sqrt{[ A^2 - y^2 ]}$

$v=\omega\times \sqrt{[ {0.06}^2 - {0.045}^2 ]}$

Velocity = 0.4762 m/s