2 physics problems

1. Apr 15, 2005

oldunion

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0 s. It then oscillates with a period of 2.05 s and a maximum speed of 58.0 cm/s.

i found the amplitude to be 18.92. they want the gliders position at time 24 seconds. my answer of 5.35 and 5.4 are both wrong.

18.92cos(2pi(24)/2.05) this is what i did and i dont know whats wrong.

Problem 2:
On your first trip to Planet X you happen to take along a 165 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 31.5 cm. You then pull the mass down 6.60 cm and release it. With the stopwatch you find that 11.0 oscillations take 15.4 s.

their question is "can you now satisfy your curiosity"? answr in m/s^2

i dont really have a clue here

2. Apr 16, 2005

learningphysics

The above equation is correct. But the numbers you gave (5.35,5.4) are wrong. Calculate the above again. Make sure you use radians on your calculator.

First find k for the spring... a spring mass oscillator's period T depends only on k and m. T = 2*pi*sqrt(k/m) You can calculate the period. You can calculate m... So you should be able to solve for k.

Now you need to find g using k.... hint: what are the forces acting on the mass when it is stretched 0.315m.... can you get another equation that lets you solve for g?

3. Apr 17, 2005

oldunion

k = .165 *(2*pi*.4756)^2 (for omega squared where omega is equal to 2piF, where f equals 10oscillations/14.5 seconds.
k=3.0981
3.0981(.315)-(.165)g=0

g=5.9147

does this look correct before i submit it and lose points.

4. Apr 18, 2005

oldunion

it is incorrect

5. Apr 18, 2005

OlderDan

You need to get some algebra correct first

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

$$k = m \left[\frac{T}{2 \pi}\right]^2$$

$$k = m \left[\frac{1}{2 \pi f}\right]^2$$

The original statement of the problem said 11 oscillations, not 10, so f = 11/14.5 seconds = . . .

Last edited: Apr 18, 2005