Simple Harmonic Motion (SHM) question check

  • #1

Homework Statement


A metal sphere of mass 0.20kg is hung from a helical spring whose top end is clamped. the sphere is displaced by 0.030m below its equilibrium position, and then released. 20 cycles of oscillation, of amplitude 0.030m, occur in a time of 12.0s.
calculate;
(I) the spring stiffness, k. (force per unit extension)
(II) the maximum velocity of the sphere as it oscillates.
(III) the maximum acceleration of the sphere as it oscillates.


Homework Equations


ω=2*∏/T

The Attempt at a Solution


my workings;
(I) W=2*∏/T
=2*∏/1.67
W= 3.76rads-1

w^2=m/k
3.76^2=0.20/k
k=0.20/3.76^2
= 0.20/14.14
k=1.41*10^-2 Nm-1

(II) Vmax= +- W √A^2 - x^2
but x=0 so
Vmax= +- WA
= +- 3.79 * 0.030
Vmax= +- 1.13*10^-1 ms-1

(III) amax= +- W^2 * x
but x=A so
amax= +- W^2 * A
= +- 3.76^2 8 0.030
amax= +- 4.24*10^-1 ms-2

could you please check my answers for me, because, i don't know why, i just feel like I've gone wrong somewhere.
thanks sarah x

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,654
Welcome to PF;

(I) Reasoning looks OK.
How did you calculate T=1.67(s?)
In 12 seconds you get 20 cycles - how much time for one cycle?

Usually better to combine equations before putting the numbers in - to avoid rounding errors.

(II) and (III) OK reasoning ... you should keep the direction (+/-) since they ask for "velocity" rather than speed.
 

Related Threads on Simple Harmonic Motion (SHM) question check

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
3
Views
11K
Replies
25
Views
635
Replies
6
Views
10K
Replies
11
Views
1K
Replies
1
Views
424
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
0
Views
3K
Replies
1
Views
4K
Top