Simple harmonic motion block question

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Homework Help Overview

The discussion revolves around a block attached to a spring undergoing simple harmonic motion. The spring constant is given as 6.7 N/m, and the amplitude of motion is 7.8 cm. The block's speed is measured at a specific position, prompting calculations for the mass of the block, the period of motion, and the maximum acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss energy conservation principles to calculate the mass of the block and question the correctness of their equations for velocity and acceleration. There is uncertainty about unit conversions and the implications of using different measurement systems (meters vs. centimeters).

Discussion Status

Some participants have provided feedback on calculations and highlighted potential errors in unit handling. There is an ongoing exploration of how to determine maximum acceleration without needing to specify the phase angle. Multiple interpretations of the equations and their applications are being discussed.

Contextual Notes

Participants are navigating through the complexities of simple harmonic motion equations and the need for consistent units in their calculations. There is a focus on ensuring clarity in the mathematical expressions used and the physical meanings behind them.

SteroidalPsyc
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Homework Statement


A block of unknown mass is attached to a spring of spring constant 6.7N/m and undergoes simple harmonic motion with an amplitude of 7.8 cm. When the mass is halfway between its equilibrium position and its max endpoint, its speed is measured to be 38.5 cm/s.
(a) Calculate the mass of the block.
(b) Find the period of the motion.
(c) Calculate the maximum acceleration.


Homework Equations


1. E = K + Us
2. E= 1/2 kA2
3. omega = \sqrt{k/m}
4. T = 2pi/w
5. x(t) = Acos(omega*t+phi)
6. v(t) = Awsin(omega]t+phi)(not sure if this is right...)
7. a(t) = -Aw2cos(wt+phi)(not really sure if I took the derivatives correctly...)


The Attempt at a Solution


I was only given the x(t) function for equations 6&7 so please correct me if I'm wrong on those. But here is what I have done so far...(lower case k is spring constant. upper case K is kinetic energy)

(a)

1/2kA2 = 1/2kx2 + 1/2mv2

[STRIKE]1/2[/STRIKE](6.7 N/m)(7.8)2 = [STRIKE]1/2[/STRIKE](6.7 N/m)(7.8/2)2 + [STRIKE]1/2[/STRIKE] m(38.5 cm/s2)

407.63 = 101.94 + 38.5 m

305.69/38.5 = m

m = 7.94 kg (should it be kg? right answer?)

(b)

w = sqrt(7.8 / 7.94) = 0.99

T = 2pi / 0.99 = 6.34 (right answer?)

(c) (this is where i really need help)

I don't know how to find amax. Do i set phi equal to a certain angle (radians)?

Here is what I have so far, but not really sure if I'm doing it right.

a(t) = -Aw2cos(wt +phi) = (7.8)(0.99)2cos{(0.99)(38.5)+phi}

I just don't know what to set phi equal to or what I need to do from here (assuming I took the derivatives of x(t) correctly.)

I really do appreciate all of your help and hard work to make this an awesome help site!

Much love,
Tim
 
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SteroidalPsyc said:

Homework Statement


A block of unknown mass is attached to a spring of spring constant 6.7N/m and undergoes simple harmonic motion with an amplitude of 7.8 cm. When the mass is halfway between its equilibrium position and its max endpoint, its speed is measured to be 38.5 cm/s.
(a) Calculate the mass of the block.
(b) Find the period of the motion.
(c) Calculate the maximum acceleration.

Homework Equations


1. E = K + Us
2. E= 1/2 kA2
3. omega = \sqrt{k/m}
4. T = 2pi/w
5. x(t) = Acos(omega*t+phi)
6. v(t) = Awsin(omega]t+phi)(not sure if this is right...)
7. a(t) = -Aw2cos(wt+phi)(not really sure if I took the derivatives correctly...)

The Attempt at a Solution


I was only given the x(t) function for equations 6&7 so please correct me if I'm wrong on those. But here is what I have done so far...(lower case k is spring constant. upper case K is kinetic energy)

(a)

1/2kA2 = 1/2kx2 + 1/2mv2

[STRIKE]1/2[/STRIKE](6.7 N/m)(7.8)2 = [STRIKE]1/2[/STRIKE](6.7 N/m)(7.8/2)2 + [STRIKE]1/2[/STRIKE] m(38.5 cm/s2)

407.63 = 101.94 + 38.5 m

305.69/38.5 = m

m = 7.94 kg (should it be kg? right answer?)
Right idea, but you dropped the exponent on v when you plugged the numbers in. Also, be careful with the units because you have both meters and centimeters floating around.
(b)

w = sqrt(7.8 / 7.94) = 0.99

T = 2pi / 0.99 = 6.34 (right answer?)
Don't forget the units.
(c) (this is where i really need help)

I don't know how to find amax. Do i set phi equal to a certain angle (radians)?

Here is what I have so far, but not really sure if I'm doing it right.

a(t) = -Aw2cos(wt +phi) = (7.8)(0.99)2cos{(0.99)(38.5)+phi}

I just don't know what to set phi equal to or what I need to do from here (assuming I took the derivatives of x(t) correctly.)
You don't need to figure out phi. All you need to consider is the range of values does cosine takes on. The max acceleration should be pretty clear then.
 
Right idea, but you dropped the exponent on v when you plugged the numbers in. Also, be careful with the units because you have both meters and centimeters floating around.

What exponent did I drop? I'm confused...

So i guess i should change everything to m?

(b)

w = sqrt(7.8 / 7.94) = 0.99

T = 2pi / 0.99 = 6.34 (right answer?)

you said not to forget units on this part. Are the units going to be N/mkg ?
 
Last edited:
SteroidalPsyc said:
What exponent did I drop? I'm confused...
Just check your work. It's a pretty obvious mistake.
So i guess i should change everything to m?
That would be safest.
you said not to forget units on this part. Are the units going to be N/mkg ?
No, those aren't the correct units. What kind of quantity is the period T?
 
Just check your work. It's a pretty obvious mistake.

So did I just forget the (-)? v(t) = -Awsin(wt)
 

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