Velocity of a mass m that moves along the x-axis

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SUMMARY

The discussion focuses on the motion of a mass \( m \) along the x-axis, governed by the equations \( \frac{dv}{dt}(mv^2 - mv_0^2) = \frac{dv}{dt}(kx_0^2 - kx^2) \) and \( \frac{1}{2}m(v^2 - v_0^2) = \frac{1}{2}k(x_0^2 - x^2) \). Participants emphasize the importance of differentiating both sides of the equations with respect to time to analyze the velocity changes. The conversation highlights the relationship between kinetic energy and potential energy in the context of classical mechanics.

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lily
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Homework Statement
A particle of constant mass m moves along the x-axis. Its velocity v and position x satisfy the equation: 1/2m(v^2 - v0^2) = 1/2k(x0^2-x^2), where k, v0 and x0 are constants. Show that whenever v does not equal 0, mdv/dt=-kx.
Relevant Equations
1/2m(v^2 - v0^2) = 1/2k(x0^2-x^2)

mdv/dt=-kx.
dv/dt(mv^2-mv0^2) = dv/dt(kx0^2-kx^2)
 
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lily said:
Relevant Equations:: 1/2m(v^2 - v0^2) = 1/2k(x0^2-x^2)
Differentiate the both hand sides by time i.e. d/dt
 
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