Oscillation, Frequency, Amplitude etc Hmwk Problem

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A 200g mass attached to a horizontal spring oscillates with a frequency of 2.0 Hz, leading to a calculated period of 0.5 seconds. The amplitude was initially miscalculated, but the correct value is 5.5 cm, derived from the relationship between velocity, angular frequency, and amplitude. The maximum speed was found to be 70 cm/s, calculated using the amplitude and angular frequency. The total mechanical energy was determined to be 0.049 J, based on the maximum speed. The discussion emphasizes the importance of correctly applying equations and understanding mechanical energy conservation in oscillatory motion.
LBRRIT2390
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Homework Statement



A 200g mass attached to a horrizontal spring oscillates at a frequency of 2.0 Hz. At one instant, the mass is at x=5.0cm and has vx = -30cm/s.

A. Determine the period. T= ____seconds

B. Determine the amplitude. A =____cm

C. Determine the maximum speed. vmax = ____ cm/s

D. Determine the total energy. ____ J

Homework Equations



A. T = 1/frequency

B. ?

C. vmax= 2(pi)(f)(A) = 2piA/T

D. ?

Others: deltaL = mg/k where k=spring constant



The Attempt at a Solution



A. Answer = 0.5 seconds from T = 1/f... T = 1/(2.0Hz)... T=0.5 secods

B.

C. vmax=2pi(2.0Hz)(A)

D.


I'm not looking for just the answers; I am in desperate need of an explanation of HOW to do the problem. Any help is greatly appreciated!
 
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Mechanical energy is conserved. What is the mechanical energy when the mass is at x=5.0 cm and has vx = -30 cm/s? That's the answer to part D.

What is an expression for the mechanical energy when the particle is at maximum displacement, i.e. when x = A? Write an equation and solve for A. This will give you the answer to part B.
 
Thank you,
I was able out to figure out the problem last night by following a post on yahoo. I want to post it step by step here for other stumped students:

Part B. Determine the amplitude.
w = 2(pi)f
w = 2(pi)(2.0Hz)
w = 12.6 rad/sec

v = w Sqrt(A2-x2)
-0.30 = (12.6rad/sec) Sqrt[ (A2) - (0.502m) ]
-0.0238 = Sqrt[ (A2) - 0.25 ]
-0.02382 = A2 - 0.25
-5.6644x10-4 = A2 - 0.25
0.2494 = A2
Sqrt (0.2494) = A
A = 0.4994m --> 5.0 cm
****** However, the correct answer from Mastering Physics is 5.5cm, where did I mess up?
A = 5.5 cm

Part C. Determine the maximum speed.
vmax = wA
vmax = (12.6rad/s)(0.055m)
vmax = 0.69m --> 70 cm

Part D. Determine the total energy.
Etotal = (mvmax2) / 2
Etotal = (0.2kg)(0.72m) / 2
Etotal = 0.049 J
 
LBRRIT2390 said:
v = w Sqrt(A2-x2)
-0.30 = (12.6rad/sec) Sqrt[ (A2) - (0.502m) ]
-0.0238 = Sqrt[ (A2) - 0.25 ]
-0.02382 = A2 - 0.25
-5.6644x10-4 = A2 - 0.25
0.2494 = A2
Sqrt (0.2494) = A
A = 0.4994m --> 5.0 cm
****** However, the correct answer from Mastering Physics is 5.5cm, where did I mess up?

When you square both sides, the left side must be positive.
 
Oh! thank you so much kuruman!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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