10A.2 SHM Mathematics

Question 1

the motion of the oscillation in SHM is a

Answer 1

projection of motion in a circle so all the circle equations work within SHM

Question 2

what is the displacement of an object in SHM

Answer 2

x = r cos (wt) derivation page 155,

Question 3

all simple harmonic oscillators can be described by

Answer 3

a sin, or cosine function which gives their displacement, velocity and acceleration over time

Question 4

the displacement of a pendulum is = to

Answer 4

x = A cos(wt) r of circle is replaced with the amp as at t = o x= A

Question 5

the restoring force over time is given as

Answer 5

f = - k A cos(wt) since x = A cos(wt)

Question 6

using the restoring force and N2L of motion the a =

Answer 6

a = - k/m A cos(wt) derivation page 155

Question 7

from a = - k/m A cos(wt) we can tell

Answer 7

that acceleration and displacement in SHM have the same form, just acceleration acts in the opposite direction, so when x is at zero so is a and when x is at max so is a

Question 8

how do we find the velocity from x = A cos(wt)

Answer 8

by differentiating x to get dx/dt which is the gradient of the graph and the gradient of a x/t graph is V so v = -Awsin(wt)

Question 9

how do we find acceleration from x = A cos(wt)

Answer 9

by differentiating x to get dx/dt which is the gradient of the graph, and the gradient of a x/t graph is V so v = -Awsin(wt), the differentiating a second time to get dv/dt or d^2x/dt^2 which dv/dt = to the gradient of a v/t graph which is equal to acceleration so a = -Aw^2cos(wt) or -w^2x since x = A cos(wt)

Question 10

if we know that mass of a oscillating object, and we know the angular velocity of an object we can find the restoring force constant k via

Answer 10

k = w^2 m (derivation page 156)

Question 11

what is the max values of v and a in SHM given that cos and sin max values are 1

Answer 11

v max = - Aw(max)

a max = -Aw^2