How to find amplitude in SHM problem

  • Thread starter Thread starter Brittany King
  • Start date Start date
  • Tags Tags
    Amplitude Shm
Click For Summary
A mass of 120g rolls down a frictionless hill, reaching a speed of 4.2 m/s before colliding with a 3.00g mass attached to a spring with a constant of 30 N/m. After the collision, the two masses stick together and enter periodic motion, with the final velocity calculated as 1.2 m/s and the angular frequency (w) as 8.45 rad/s. There is a discrepancy in the amplitude calculation, with one participant suggesting A should be 0.142 m while the correct answer is 0.266 m. The motion can be described by the equation x(t) = 0.266sin(8.44t). The discussion highlights the importance of correctly identifying mass values and understanding the mechanics of the system.
Brittany King
Messages
11
Reaction score
0

Homework Statement


A mass of 120g rolls down a frictionless hill reaching a speed of 4.2 m/s and collides with another mass 3.00g attached to a spring of constant 30 N/m. The two masses stick together and enter into periodic motion. What is the equation for the motion?

Homework Equations


m1v1+m2v2=mtV
x(t)=Acos(wt+phi)
w=k/m^1/2
Vmax=wA

The Attempt at a Solution


I found Vfinal=1.2 m/s
Also found w=8.45 rad by w=k/m^1/2
I thought A would be 0.142 but in the answer A=0.266
the answer is: x(t)=0.266sin(8.44t)
 
Physics news on Phys.org
I guess you meant to type 300 g rather than 3.00 g for the second mass.

Your work looks good to me. I agree that A = 0.142 m.

[EDIT: I'm assuming that the first mass reaches the bottom of the hill and then moves horizontally before hitting the spring as shown below.]

(If there is no friction on the hill, the 120 g mass would slide rather than roll down the hill.)
 

Attachments

  • pic.png
    pic.png
    1.5 KB · Views: 558
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
967
  • · Replies 1 ·
Replies
1
Views
2K
Replies
26
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K