# Pendulum and Simple Harmonic Motion Problems!

1. Jan 19, 2009

### think4432

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 dont understand the spring constant equation.

2. Jan 19, 2009

### rock.freak667

1) You know that the period of a pendulum is
$$T = 2\pi \sqrt{\frac{l}{g}}$$

and that $f= \frac{1}{T}$

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. Jan 19, 2009

### think4432

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. Jan 19, 2009

### rock.freak667

Yes solve for a. But recheck your rearranging of terms.

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

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:

$$T= 2\pi \sqrt{\frac{m}{k}}$$

5. Jan 19, 2009

### think4432

thank you so much!

great help!

now i might pass the test tommorow!

thanks again!

6. Feb 24, 2009

### skiing4free

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