Simple harmonic motion block question

In summary: Awsin(wt)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?T is the time right? So the units are just s?In summary, a block with an unknown mass attached to a spring with a spring constant of 6.7N/m undergoes simple harmonic motion with an amplitude of 7.8 cm and a speed of 38.5 cm/s when it is halfway between its equilibrium position and its max endpoint. To find the mass, the kinetic energy and velocity equations were used, resulting in a mass of 7.94 kg. The
  • #1
SteroidalPsyc
6
0

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 = [tex]\sqrt{k/m}[/tex]
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
 
Physics news on Phys.org
  • #2
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 = [tex]\sqrt{k/m}[/tex]
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.
 
  • #3
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:
  • #4
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?
 
  • #5
Just check your work. It's a pretty obvious mistake.

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

1. What is simple harmonic motion?

Simple harmonic motion refers to the type of motion in which an object moves back and forth in a regular and repeating manner. It is characterized by a restoring force that is directly proportional to the displacement of the object from its equilibrium position. Examples of simple harmonic motion include a mass on a spring and a pendulum.

2. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A cos(ωt + φ), where x is the displacement of the object, A is the amplitude, ω is the angular frequency, and φ is the phase angle. This equation describes the position of the object at any given time during its motion.

3. What is the difference between simple harmonic motion and oscillatory motion?

Simple harmonic motion is a specific type of oscillatory motion in which the restoring force is directly proportional to the displacement of the object. Other types of oscillatory motion may have different relationships between the force and displacement, such as in damped or forced oscillations.

4. What are the applications of simple harmonic motion?

Simple harmonic motion has a wide range of applications in science and engineering. It is used in clocks and watches to keep time, in musical instruments to produce sound, in seismology to study earthquakes, and in many other fields. Simple harmonic motion is also an important concept in understanding more complex systems, such as in wave mechanics and quantum mechanics.

5. How does amplitude affect simple harmonic motion?

The amplitude of simple harmonic motion is the maximum displacement of the object from its equilibrium position. A larger amplitude results in a greater displacement and a longer period of oscillation. The amplitude also affects the maximum potential and kinetic energy of the object, as well as the maximum velocity. In general, a larger amplitude results in a more energetic and faster motion.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
396
  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
864
  • Introductory Physics Homework Help
Replies
10
Views
913
  • Introductory Physics Homework Help
Replies
5
Views
836
Replies
13
Views
310
  • Introductory Physics Homework Help
Replies
1
Views
844
Back
Top