2. Given the following position-time graph of a simple harmonic oscillator, at what
times will the velocity be at its maximum? What times will the acceleration be at
its maximum?
A) Maximum velocity at t=0.5 + 0.5T, maximum acceleration at t=0 + 0.5T
B) Maximum velocity at t=0.5 + T, maximum acceleration at t=0+T
C) Maximum velocity at t=0 + 0.5T, maximum acceleration at t=0.5+0.5T
D) Maximum velocity at t = 0 + T, maximum acceleration at t=0.5+T
E) Both the maximum velocity and acceleration occur at t=0+0.5T
Note: t = time, T = period
http://www.unistudyguides.com/wiki/File:SHM_2.jpg
3. Solution: A) Maximum velocity at t=0.5 + 0.5T, maximum acceleration at t=0 + 0.5T
One can think about the conservation of
energy. When the position is 0m, the potential
energy is also 0 and so the only energy present
is kinetic energy – resulting in a maximum
velocity at this point. This repeats for every half
period (answer B is not fully correct as it only
accounts for either the +v or –v; it should
include both).
The maximum acceleration occurs when the
displacement is also at its maximum values.
“For a motion to be simple harmonic motion,
the acceleration of the object must be
proportional to its displacement from the
equilibrium position and opposite in direction,
at all times.”
http://www.unistudyguides.com/wiki/File:SHM_2.jpg
4. Answer C is incorrect because at t=0+0.5T, the displacement
is at its maximum. Again with conservation of energy, the total
energy at that point is equal to only the potential energy (no
kinetic energy). For this reason, the velocity would be 0m/s at
such times.
Answer D is incorrect, following the same reason explained for
answer C.
Answer E is incorrect. Although the description for
acceleration is correct, the times for velocity is incorrect (see
reasoning for answer C).
Explanation continued…
