Physics simple harmonic oscillations

amy1999

Part of a building structure is oscillating with approximately simple harmonic motion with a period of 10 s and amplitude 1 m.
a) calculate the maximum speed of the oscillation.
b) calculate the maximum acceleration of the oscillation.
Can you explain how to work these out please as I have no clue, the equations are confusing me.

Reply 1

TSR Jessica

19

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

Just quoting in Fox Corner so she can move the thread if needed :wizard:

Spoiler

Reply 2

Fox Corner

21

popped this in the physics forum for you

Reply 3

Joinedup

20

The relevant equations appear in the formula book as

v_max=2πfA
a_max=(2πf)² A

A (capital) is the amplitude
f is the frequency of oscillation... which is 1/T (T is the period in seconds)

Original post by amy1999

Part of a building structure is oscillating with approximately simple harmonic motion with a period of 10 s and amplitude 1 m.
a) calculate the maximum speed of the oscillation.
b) calculate the maximum acceleration of the oscillation.
Can you explain how to work these out please as I have no clue, the equations are confusing me.

In simple harmonic motion, the acceleration is always directly proportional to displacement and has the following defining equation:

a = - \omega^2 x

----eqn(1)

where

a

is the acceleration,

x

is the displacement and

\omega

is the angular frequency , given by

\omega = 2 \pi f = \frac{2 \pi}{T}

.

From eqn(1), the maximum value of acceleration occurs at when the displacement is maximum - maximum displacement implies amplitude, A.

So the maximum acceleration is

a = \omega^2 A = (2 \pi f)^2 A

----given by Joinedup

The displacement of simple harmonic motion can be described by a sinusoidal function

x = A \cos( \omega t)

----eqn(2)

If equation (2) is differentiated with respect to time t, we obtain the "velocity"

v = -\omega A\sin( \omega t)

----eqn(3)

The maximum speed would occur when

\sin( \omega t) = \pm 1

, so we have

v = \omega A = (2 \pi f)A

----given by Joinedup

Reply 5

swinroy

12

Original post by amy1999

Part of a building structure is oscillating with approximately simple harmonic motion with a period of 10 s and amplitude 1 m.
a) calculate the maximum speed of the oscillation.
b) calculate the maximum acceleration of the oscillation.
Can you explain how to work these out please as I have no clue, the equations are confusing me.

The max speed is the angular speed x Amplitude
The angular speed is calculated as follows: 2 x Pi divided by Period

THe max acceleration is the previous answer x angular speed

if you liken circular motion to SHM
one revolution is equivalent to one complete oscillation
in both cases 2 Pi RADIANS are performed in a periodic time
The angular speed is the number or RADIANS performed per SECOND

Physics simple harmonic oscillations

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