Simple harmonic motion of a machine part

In summary, the machine part is undergoing simple harmonic motion with a frequency of 5hz and an amplitude of 1.80cm. To find the time it takes for the part to go from x = 0 to -1.80cm, you can manipulate the equation x = Acoswt and use the fact that the displacement is the negative of the amplitude. By marking the points 1.8cm and -1.8cm on a displacement vs time graph and using the period (T = 1/5), it can be determined that the time is either 0.05s or 0.25s. Both answers are valid.
  • #1
kaywond
2
0

Homework Statement


A machine part is undergoing SHM with a frequency of of 5hz and amplitude of 1.80cm. How long does it take the part to go from x = 0 to -1.80cm?


Homework Equations


x = Acoswt


The Attempt at a Solution


X is given and convert it to metres 0.018. I need to manipulate answer to get time.
angular velocity is 2pi x 5 = 31.41. It is just an equation manipulation. could someone please show me the technique?
 
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  • #2
The displacement is the negative of the amplitude.
 
  • #3
If the peak amplitude is 1.8cm then asking when the displacement is -1.8 makes it rather easy. No need to even use the sin button on the calculator :-)

Sketch displacement vs time. Mark the points 1.8m and -1.8m. If T is the period (=1/5) at what fraction of T does 1.8cm and -1.8cm occur?
 
  • #4
thanks 1/5 equals 0.2. Then divide by 4 and 0.05 is the time in seconds
 
  • #5
0.05s is one valid answer but there are two possible valid right answers for this question. I recommend you give both.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object moves back and forth in a regular pattern. It occurs when a restoring force is applied to an object that is proportional to its displacement from its equilibrium position.

2. What is an example of simple harmonic motion in a machine part?

An example of simple harmonic motion in a machine part is the motion of a spring in a car's suspension system. As the car drives over bumps in the road, the spring is compressed and then released, causing it to oscillate back and forth in a regular pattern.

3. How is the period of simple harmonic motion calculated?

The period of simple harmonic motion is calculated using the equation T = 2π√(m/k), where T is the period in seconds, m is the mass of the object in kilograms, and k is the spring constant in newtons per meter. This equation shows that the period is dependent on the mass and stiffness of the object.

4. What factors can affect the amplitude of simple harmonic motion?

The amplitude of simple harmonic motion can be affected by the initial displacement of the object from its equilibrium position, the strength of the restoring force, and any external forces acting on the object. The amplitude can also decrease over time due to friction and other forms of energy loss.

5. How is simple harmonic motion used in machines?

Simple harmonic motion is used in machines to create controlled and predictable movements. It is commonly used in pendulums, springs, and gears to convert energy into mechanical motion. This type of motion is also used in clocks, washing machines, and car suspensions to maintain smooth and consistent movements.

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