Simple harmonic motion block question

AI Thread Summary
A block attached to a spring with a spring constant of 6.7 N/m undergoes simple harmonic motion with an amplitude of 7.8 cm and a measured speed of 38.5 cm/s at the halfway point. The mass of the block can be calculated using energy conservation principles, yielding approximately 7.94 kg, though unit consistency is crucial. The period of the motion is derived from the angular frequency, resulting in a value of 6.34 seconds, but attention to units is necessary. Maximum acceleration can be determined without needing to define the phase angle, focusing instead on the maximum values of the cosine function. The discussion emphasizes the importance of unit conversion and careful calculation in solving harmonic motion problems.
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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