Physics hacksaw problem, harmonic motion

[rlc]

Homework Statement

One end of a light hacksaw blade is clamped in a vise with the long axis of the blade horizontal and with the sides vertical. A 0.665- kg mass is attached to the free end. When a steady sideways force of 20.5 N is applied to the mass it moves aside 13.3 cm from its equilibrium position. When released, the blade moves back and forth with harmonic motion. Approximately how many complete oscillations will the blade make in 30.0 s when the force is removed?

Homework Equations

(I don't know any...)

The Attempt at a Solution

I have absolutely no clue where to begin with this problem. Please help!

 

[Dr.D]

The saw blade and attached mass form a spring-mass system. The information about the static deflection is intended to enable you to gain information about the effective stiffness. When the system is released, it will oscillate at its natural frequency (which you should be able to find), so you can determine the number of cycles that will occur in 30 sec.

 

[rlc]

frequency=velocity/wavelength
Velocity=displacement/time 0.133m/30s=0.00443333
wavelength=speed/frequency

 

[rlc]

Assuming k = 20.5N / 0.133m = 154.14 N/m

For single harmonic motion... ω² = k/m
ω = √(154.14 / 0.665) = 15.2246 rad/s

f = ω/2π = 15.2246/(2π) ... f ≈ 2.423 Hz
Cycles in 30s = 2.423 sˉ¹ x 30s .. .. .. ►≈ 72.7 oscillations

But this isn't right. Where did I go wrong?

 

[OldEngr63]

How do you know this is wrong?

 

[rlc]

I'm doing the homework through a program called LONCAPA. It's through my university, and I can submit my answers (I only have 7 tries) and it tells me if it is right or wrong, depending on what was put into it for right vs. wrong. I submitted this and it told me it was wrong.

 

[Dr.D]

I'm afraid you are caught between a rock and a hard place. I don't see anything wrong with your solution once you got into it.

 

[rlc]

Maximum speed=(displacement)SQRT(k/mass)
Max speed=(0.133 m)SQRT((20.5/0.133)/0.665)=2.02484 m/s

Period=(2*pi*displacement)/max speed
Period=(2*pi*0.133m)/(2.02484 m/s)=0.4127

Frequency=1/period
Frequency=1/0.4127
Frequency=2.423039 Hz (WHICH IS THE SAME NUMBER I GOT BEFORE) :(

 

[rlc]

Question, does the fact that the question asks "when the force is removed" change this problem?

 

[Dr.D]

No, that simply means that the system is released to oscillate.

But here are some words that may make a difference: "Approximately how many complete oscillations will the blade make in 30.0 s when the force is removed?" I think the answer is required to be an integer.

 

[rlc]

Makes sense. But how do you change it to an integer?

 

[rlc]

Wow, I'm dumb.

 

[rlc]

The answer was 72 oscillations, no decimal.

Thank you so much!