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

Click For Summary
The discussion centers on the relationship between velocity and displacement for a mass m moving along the x-axis, using the equation dv/dt(mv^2 - mv0^2) = dv/dt(kx0^2 - kx^2). Participants explore the implications of differentiating both sides with respect to time, emphasizing the connection between kinetic and potential energy. The relevant equation 1/2m(v^2 - v0^2) = 1/2k(x0^2 - x^2) is highlighted to illustrate energy conservation. The conversation delves into the mathematical derivation and physical interpretation of these relationships. Overall, the discussion aims to clarify the dynamics of motion under the influence of spring forces.
lily
Messages
1
Reaction score
0
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)
 
Physics news on Phys.org
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 27 ·
Replies
27
Views
879
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
671
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K