- #1
vf_one
- 14
- 0
Hi everyone
I'm having trouble using the equations for SHM calculations. We got told to set them on radians but I forgot to when I was doing the problems and got some of the answers right. But when I tried to do them when the calculator was set in radians, the answers were totally wrong.
We are given a set of equations to use
x=Xcos(2πft)
v= -X√(k/m)sin(2πft)
a= -XK/mcos(2πft)
Do we set the calculator to radians when we use these equations?
Do we take into account the negative "-" sign in front of X for the velocity and acceleration equations?
I'll show you the working a question I did and maybe you can see where I went wrong.
A 64kg bungy jumper is displaced 13m from equilibrium downwards and then released. She bounces up and down with a period of 5.4s. When she is 5m abover her initial rest position and moving upwards, what is her velocity?
2. The attempt at a solution
I calculated this with my calculator set in degrees
x=Xcos(2πft)
5=13cos(2π x 1/5.4t)
t= 57.9s
v= -X√(k/m)sin(2πft)
v= -13√(86.6/64)sin(2π x 1/5.4 x 57.9)
v= -14m/s
The answer was 14m/s. Why was the negative sign ignored?
But when we were asked to calculate the acceleration at 5m displacement the answer is
-6.77m/s2
a= -XK/mcos(2πft)
a= -13 x 86.6/64cos(2π x 1/5.4 x 57.9)
a= -6.77m/s2 Calculator still set in degrees
If anyone could help that would be very appreciated. Thanks
I'm having trouble using the equations for SHM calculations. We got told to set them on radians but I forgot to when I was doing the problems and got some of the answers right. But when I tried to do them when the calculator was set in radians, the answers were totally wrong.
We are given a set of equations to use
x=Xcos(2πft)
v= -X√(k/m)sin(2πft)
a= -XK/mcos(2πft)
Do we set the calculator to radians when we use these equations?
Do we take into account the negative "-" sign in front of X for the velocity and acceleration equations?
I'll show you the working a question I did and maybe you can see where I went wrong.
Homework Statement
A 64kg bungy jumper is displaced 13m from equilibrium downwards and then released. She bounces up and down with a period of 5.4s. When she is 5m abover her initial rest position and moving upwards, what is her velocity?
2. The attempt at a solution
I calculated this with my calculator set in degrees
x=Xcos(2πft)
5=13cos(2π x 1/5.4t)
t= 57.9s
v= -X√(k/m)sin(2πft)
v= -13√(86.6/64)sin(2π x 1/5.4 x 57.9)
v= -14m/s
The answer was 14m/s. Why was the negative sign ignored?
But when we were asked to calculate the acceleration at 5m displacement the answer is
-6.77m/s2
a= -XK/mcos(2πft)
a= -13 x 86.6/64cos(2π x 1/5.4 x 57.9)
a= -6.77m/s2 Calculator still set in degrees
If anyone could help that would be very appreciated. Thanks