A 0.540 kg glider on an air track is attached to the end of an ideal spring with

Question

A 0.540 kg glider on an air track is attached to the end of an ideal spring with force constant 457 N/m ; it undergoes simple harmonic motion with an amplitude of 4.10×102 m . Part A Compute the maximum speed of the glider. vmax = m m/s SubmitMy AnswersGive Up Part B Compute the speed of the glider when it is at x = 1.30×102 m . v = m m/s SubmitMy AnswersGive Up Part C Compute the magnitude of the maximum acceleration of the glider. amax = m m/s2 SubmitMy AnswersGive Up Part D Compute the acceleration of the glider at x = 1.30×102 m . a = m m/s2 SubmitMy AnswersGive Up Part E Compute the total mechanical energy of the glider at any point in its motion. Etotal = m J

Explanation / Answer

The centre of the SHM is at x = 0, the rest positions are at x = 4.10*10-2,

the maximum speed is at x = 0.
(a) The total mechanical energy of the glider is constant throughout the SHM.

This energy is 0.540kx² + 0.540mv² = 0.540 x 457 x (4.10*10-2)² + 0.5 x 0.5 x 0²

(taking the position where x = 4.10*10-2, v = 0)

= 0.540 x 457 x 0² + 0.540 x 0.540 x v² (taking a position where x = 0).

This gives 0.2916v² = 0.540 x 457 x (4.10*10-2)2

v = 2.036m/s.
(b) We have 0.540 x 457 x (1.30×102)² + 0.540 x 0.540v² = 0.540x 457 x (4.10*10-2)2

v = 1.13m/s.
(c) This occurs when the spring force on the glider, kx, is at a maximum ie when x = ±4.10*10-2. Magnitude of maximum acceleration is then |kx|/m = (457 x4.10*10-2)/0.540 = 41.53m/s².
(d) This is kx/m = 450 x (1.30×102)/0.540 = -10.83m/s² (the acceleration is towards the centre of the oscillation).
(e) This is (by part (a)) 0.540 x 457 x(4.10*10-2)² = 0.414J.

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