What is the relationship between simple harmonic motion and amplitude?

[dustybray]

Hi,

I'm having trouble with this problem because I don't know how to deal with all these unknowns:

4. A block attached to a spring with unknown spring constant oscillates with a period of 2.0 s. What is the period if :

a. The mass is doubled?
b. The mass is halved?
c. The amplitude is doubled?
d. The spring constant is doubled?

I think I should use T = 2π * sqrt( m / k ), but what are m and k... ??

Also, what relates this to amplitude?


Hopefully I'm doing this problem correctly, but I don't know how to get acceleration:

7. The bow of a destroyer undergoes simple harmonic vertical pitching motion with a period of 8.0 s and an amplitude of 2.0 m.

a. What is the maximum vertical velocity of the destroyer’s bow?

f = 1/T = .125rev/s

ω = 2πf = .785rad/s

v = ωr = (.785rad/s) (2m) = 1.57m/s

b. What is the maximum acceleration?

??

c. An 80 kg sailor is standing on a scale in the bunkroom in the bow. What are the maximum and minimum readings on the scale in Newtons?

80kg * ( 9.8m/s^2 + a)

80kg * ( 9.8m/s^2 - a)


Thanks,

dusty...

 

[BerryBoy]

m = the mass, k = spring constant

You will have to write your answers in terms of 'T'.

Also, what relates this to amplitude?

Well amplitude isn't in the equation for the period, so does it really affect T?

I have to go to a lecture now... I'll finsh this reply when I return.
(Quickly: if x = A Sin( wt),

Then - dx/dt = v = Aw Cos(wt) - You noticed that v is a maximun at Cos(wt) = 1. (You used 'r' for amplitude, whereas I use 'A').

So dv/dt = a. Hope this helps,

Regards,
Sam

 

[dustybray]

Ok,

I'm not sure if I understood, but I've reworked the problems:

For this problem, I hope I used logical mathematics and not contrived black magic...

4. A block attached to a spring with unknown spring constant oscillates with a period of
2.0 s. What is the period if :

T = 2π * sqrt( m / k )

a. The mass is doubled?

T = 2π * sqrt( 2 * (m / k) )

T = sqrt( 2 ) ( 2π * sqrt( m / k ) )

T = sqrt( 2 ) * 2s

T = 2.83s

b. The mass is halved?

T = sqrt( .5 ) * 2s

T = 1.41s

c. The amplitude is doubled?

2s – not dependant on amplitude

d. The spring constant is doubled?

T = sqrt( .5 ) * 2s

T = 1.41s


So, do you mean that a = v[max] / period ?

7. The bow of a destroyer undergoes simple harmonic vertical pitching motion with a period of 8.0 s and an amplitude of 2.0 m.

a. What is the maximum vertical velocity of the destroyer’s bow?

f = 1/T = .125rev/s

ω = 2πf = .785rad/s

v = ωr = (.785rad/s) (2m) = 1.57m/s

b. What is the maximum acceleration?

a = Δv / Δt

a = (1.57m/s) / (8s) = .2m/s^2

c. An 80 kg sailor is standing on a scale in the bunkroom in the bow. What are the
maximum and minimum readings on the scale in Newtons?

F = ma


F[max] = 80kg * ( 9.8m/s^2 + .2m/s^2 ) = 800N

F[min] = 80kg * ( 9.8m/s^2 - .2m/s^2 ) = 768N


And here is an aditional problem which I've worked, if you don't mind checking that I'm on the right path.

6. The position of a particle is given by 0.07 cos (6п t) m, where t is in s.

a. What are the frequency and the period?

ω = 2πf = 6π

f = ω / 2π = 6π / 2π

f = 3rev/s


T = 1/f = .33s

b. What is the amplitude?

.07m

c. What is the maximum speed?

v = rω = (.07m) (6π rad/s)

v = .0037m/s

d. What is the maximum acceleration?

a = Δv / Δt

a = (.0037m/s) / (.33s) = .0112m/s^2

e. What is the first time after t = 0 that the particle is at the equilibrium position?

cos (6π*t) = 0

6π*t = cos^-1 ( 0 )

t = cos^-1 ( 0 ) / 6π

t = .083s

f. What is the first time the particle is at x = 0 and moving to the right?

cos (6π*t) = 0

6π*t = cos^-1 ( 0 ) = (1/2)π, but going left

(1/2)π + π = (3/2)π

t = (3/2)π / 6π

t = .25s

Thanks for all the help,

dusty...

 

[BerryBoy]

I agree with all of your answers to question 4.

I'm sorry, I seem to have confused you with Question 7, let me try again...

We can say the the position x for an object undergoing SHM can be written as:

x = A Sin (wt)

By definition, if we differentiate a function of displacement with respect to time (dx/dt), we get velocity. So...

v = dx/dt = d/dt(A Sin wt) = Aw Cos wt

Also by definition if we diferentiate a function of velocity with respect to time (dv/dt), we get acceleration. So can you now differentiate:

d/dt(Aw Cos (wt)) If you can't don't worry... I'm just trying to boost your knowledge of SHM. If you understand this, then great! If not, let me tell you that:

a_max = Aw² (this equation is given in exams that I have taken in sixth-form and University).

Let me know how you get on,
Sam