Pendulum and Simple Harmonic Motion Problems

In summary, on planet X, the acceleration due to gravity is .70 hz and the period of a 200 gram mass vibrating on a spring is 4.80 seconds.
  • #1
think4432
62
0
1. Determine the value of the acceleration due to gravity on planet X if a pendulum having a length of 30.0 cm has a frequency of .70 hz on that planet?

2. Find the period of motion for a 200 gram mass vibrating on a spring having a spring constant of 4.0 n/m?

3. When a mass of 35.0 grams is attached to a certain spring, it makes 20 complete vibrations in 4.80 seconds. What is the spring constant for this spring?


equations: SHM: T=2pi sqr(m/k)


1. i have no idea...really
2. t=2pi sqr(200g/4 n/m) = 44.4288
3. k=4pi^2m/t^2
k=4pi^2(35g)/(4.1s)^2 [t=4.1s]
k=82.19 N/M


I don't understand the spring constant equation.

Please help?
 
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  • #2
1) You know that the period of a pendulum is
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

and that [itex]f= \frac{1}{T}[/itex]

2)Should be correct since you are using the correct formula

3) The periodic time, T is the time taken for one vibration.

20 vibrations occur in 4.8 s
1 vibration will occur in how much time?

Now use the equation T=2pi sqr(m/k) and rearrange it to get k
 
  • #3
Ohh! Ok! I understand number 2, and 3.

But number 1 is still a little confusing.

So, T=2pi sqrt(L/a)
Frequency, f=1/T
f=(1/2pi)sqrt(a/L)

solve for a?

a=(1/4pi^2)L/T^2
(1/4pi^2)(30cm)/1.42^2

a=36.70 m/s^2 ?



and one more problem:
A 70.0 gram mass is hung from a vertical spring causing it to stretch 20.0 cm. Find the period of motion for this system when the 70.0 gram mass is replaced by a 90.0 gram mass and set into simple harmonic motion.

how would i set that up?

thanks for the help on the other 2 problems! :]
 
  • #4
think4432 said:
Ohh! Ok! I understand number 2, and 3.

But number 1 is still a little confusing.

So, T=2pi sqrt(L/a)
Frequency, f=1/T
f=(1/2pi)sqrt(a/L)

solve for a?

a=(1/4pi^2)L/T^2
(1/4pi^2)(30cm)/1.42^2

a=36.70 m/s^2 ?
Yes solve for a. But recheck your rearranging of terms.

f = 1/2pi sqrt(a/l)
f2= (1/4pi2)(a/l)

think4432 said:
and one more problem:
A 70.0 gram mass is hung from a vertical spring causing it to stretch 20.0 cm. Find the period of motion for this system when the 70.0 gram mass is replaced by a 90.0 gram mass and set into simple harmonic motion.

how would i set that up?

When you placed the 70g mass (the weight causes it to stretch) on the spring it stretched by 20 cm, as it said right?
Remember Hooke's law? You can now find the spring constant,k.

When you have k, to find the period when the mass is 90grams just use the formula you stated above:

[tex]T= 2\pi \sqrt{\frac{m}{k}}[/tex]
 
  • #5
thank you so much!

great help!

now i might pass the test tommorow!

thanks again!
 
  • #6
think4432 said:
2. t=2pi sqr(200g/4 n/m) = 44.4288

! 200g=.2kg! ALWAYS CONVERT, you CANNOT multiply grams by meters it WONT WORK!
 

1. What is a pendulum and how does it work?

A pendulum is a weight attached to a fixed point by a string or rod, which is able to swing back and forth under the influence of gravity. This motion is known as simple harmonic motion, where the pendulum swings in a regular pattern with a constant period.

2. What factors affect the period of a pendulum?

The period of a pendulum is affected by the length of the string or rod, the weight of the pendulum, and the strength of gravity. The longer the string, the longer the period. A heavier pendulum will also have a longer period. Gravity plays a role in the period, as a stronger gravitational force will result in a shorter period.

3. How is simple harmonic motion related to a pendulum?

Simple harmonic motion is the back and forth motion of a pendulum. This motion is caused by the force of gravity pulling on the pendulum as it swings back and forth. The restoring force of gravity causes the pendulum to return to its original position, creating a regular and predictable motion.

4. What is the equation for the period of a pendulum?

The equation for the period of a pendulum is T = 2π√(L/g), where T is the period, L is the length of the string or rod, and g is the acceleration due to gravity. This equation shows that the period is directly proportional to the length of the pendulum and inversely proportional to the strength of gravity.

5. How can pendulum and simple harmonic motion problems be applied in real life?

Pendulums and simple harmonic motion have many practical applications, such as in clocks, metronomes, and seismometers. They are also used in engineering and physics to study oscillations and vibrations in various systems. Understanding the principles of pendulum and simple harmonic motion can also help in designing and analyzing structures, such as bridges and buildings, to ensure their stability and safety.

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