The amplitude remains constant
Calculate the magnitude of the maximum acceleration when the time period is 0.2s and the maximum displacement is 0.02m.
Use the equation a=-ω^2 * xo
a=(2π/0.2)^2 * 0.02=19.7m/s
Calculate the speed of the oscillation when the time taken is 0.12s to displace by 1.62cm and maximum displacement is 0.02m.
Use the equation v = (2π/t)√(xo^2-x^2) Answer
v = (2π/t)√(xo^2-x^2)
v = (2π/0.12)√((0.02)^2-(0.0162^2)
v = 31.4√((0.02)^2-(0.0162^2) = 0.37m/s
Following the question above, if the diplacement was in a negative direction would be the direction of its motion.
Even though it would seem logical that the motion would also be in the negative direction it is actually positive. This is because if it were displaced by -0.0162 m, -0.0162^2 is actually a positive number, therefore making the velocity positive. As velocity is a vector quantity, if it is positive it shows that the direction of motion is to the right in IB physics.
End of this part! Please read next part as well!
Drafted by Gina