Block- spring- bullet- harmonic motion

  • #1
~christina~
Gold Member
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[SOLVED] Block- spring- bullet- harmonic motion...

Homework Statement



Your assignment is to make measurements of the muzzle velocity of a bullet. This measurig device consists of a 10kg block which rests on a horizontal surface of negligible mass and is attatched to a hooke's law spring. The other end of the springis fixed to the wall. The preliminary data sent to you shows a 9.72g bullet traveling horizontally and colliding with the block, and as a result of this impact the bullet -block combination compresses the spring 7.50cm. This bullet-block system then executes simple harmonic motion with a 0.775Hz frequency of oscillation.

a) Give the equation which describes the displacement of this system as a function of time
b) What is the value of the spring constant?

c) What is the max acceleration of the system

d) What is the value of the displacement at this acceleration?

e) What is the total energy of this system?

f) what is the muzzle velocity of the gun that fired the bullet?


Homework Equations


v= omega/k
omega= 2pi*f
momentum equation for inelastic collision = > m1v1 + m2v2= (m1+ m2) vf
F= ma
F= -kx

The Attempt at a Solution



a) Give the equation which describes the displacement of this system as a function of time

[tex]y(x,t)= A sin (\omega*t -kx)[/tex]

[tex]\omega= 2 \pi f= (0.775Hz)2\pi= 4.869 rad/s[/tex]

[tex]A= 7.50cm=> 0.0750m [/tex]

k=?


I need help Thanks.
 
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Answers and Replies

  • #2
Doc Al
Mentor
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b) What is the value of the spring constant?

[tex]T= 2\pi \sqrt{ \frac{k} {M+m} [/tex]

my lab teacher said that

[tex]k= \frac {m+M} {f^2 4 \pi^2} [/tex] but I don't know how he got this...
He just used T = 1/f and then solved for k.
 
  • #3
kamerling
454
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[tex]
y(x,t)= A sin (\omega*t -kx)
[/tex]

This is the answer to a wave equation, which isn't needed here. What is needed is just [tex]y(t)= A sin(\omega*t)[/tex].

To get k write down the second order differential equation for the harmonic motion, and
substitute [tex] A sin(\omega*t)[/tex] in it.
 
  • #4
~christina~
Gold Member
717
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This is the answer to a wave equation, which isn't needed here. What is needed is just [tex]y(t)= A sin(\omega*t)[/tex].

To get k write down the second order differential equation for the harmonic motion, and
substitute [tex] A sin(\omega*t)[/tex] in it.

it isn't? ...well my teacher gave the class that one for some reason..

is it this:
[tex]y(t)= A sin(\omega*t)[/tex]

[tex]y'(t)= \omega A cos(\omega*t) [/tex]
[tex]y''(t)= - \omega^2 A sin(\omega*t) [/tex]

then
[tex] A sin(\omega*t)= - \omega^2 A sin(\omega*t) [/tex]

how does this help?

He just used T = 1/f and then solved for k.

Oh..I was trying to manipulate it to equal that but couldn't quite figure that out.

And I added more stuff to the original post. Not sure if it's correct though

Thank you
 
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  • #5
~christina~
Gold Member
717
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b) What is the value of the spring constant?

[tex]T= 2\pi \sqrt{ \frac{k} {M+m} [/tex]

[tex]k= \frac {m+M} {f^2 4 \pi^2} [/tex] but I don't know how he got this...


[tex] k= \frac{10.00kg + 0.00972kg} {(0.775Hz)^2 4 \pi^2)} [/tex]

[tex] k= 0.42214 [/tex]

c) max acceleration of system
F= ma
F= -kx

(m+ M)a= kx

so

[tex]a= \frac{0.42214x0.075m} {(0.00972kg + 10.00kg)} = 0.0031629m/s^2 [/tex]

d) value of displacement at that acceleration

value is same as A so it's 0.075m (same given in question)

e) total E of system

[tex]E= 0.5 k\Delta A^2= 0.001187J[/tex] => that seems a bit small though

This is what I did ...I tried to do the rest of the problem besides a) (not sure about that)and I'm not sure it's correct or not.

Can someone help me out with checking this?

Thanks
 
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  • #6
kamerling
454
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is it this:
[tex]y(t)= A sin(\omega*t)[/tex]

[tex]y'(t)= \omega A cos(\omega*t) [/tex]
[tex]y''(t)= - \omega^2 A sin(\omega*t) [/tex]

then
[tex] A sin(\omega*t)= - \omega^2 A sin(\omega*t) [/tex]

how does this help?

The force of the spring on the block is -ky, so y''(t) = -ky/(m+M) (using F=ma).
now if you substitute [tex]y(t)= A sin(\omega*t)[/tex] in that, you'll find a
relation between k and [tex]\omega[/tex]
 
  • #7
kamerling
454
0
b) What is the value of the spring constant?

[tex]T= 2\pi \sqrt{ \frac{k} {M+m} [/tex]

[tex]k= \frac {m+M} {f^2 4 \pi^2} [/tex] but I don't know how he got this...

These are actually wrong. It should be [tex]T= \sqrt{ \frac{M+m} {4 \pi^2 k}[/tex]
using T=1/f and solving for k can get you the right second formula.
It's probably more worthwile to solve the differential equation yourself, and work
only with [tex]\omega[/tex]
 
  • #8
~christina~
Gold Member
717
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The force of the spring on the block is -ky, so y''(t) = -ky/(m+M) (using F=ma).
now if you substitute [tex]y(t)= A sin(\omega*t)[/tex] in that, you'll find a
relation between k and [tex]\omega[/tex]

it's [tex]\omega= \sqrt (k/m) [/tex]
 
  • #9
~christina~
Gold Member
717
0
These are actually wrong. It should be [tex]T= \sqrt{ \frac{M+m} {4 \pi^2 k}[/tex]
using T=1/f and solving for k can get you the right second formula.
It's probably more worthwile to solve the differential equation yourself, and work
only with [tex]\omega[/tex]

so based on yours I got

[tex]T= \sqrt (\frac{M+m} {4\pi^2 k}) [/tex]

using T= 1/f

[tex](1/f)^2= \frac{M+m} {4\pi^2 k} [/tex]
[tex] 4\pi^2 k= \frac{M+m} {1/f^2} [/tex]
[tex] k= \frac{f^2(M+m)} {4\pi^2} [/tex]

I think this is correct now.
 
  • #10
kamerling
454
0
And since you know [tex]\omega[/tex] you can now compute k, and the rest shouldn't give any more problems. The energy in e) will still seem small, but that's because the inelastic collision between the bullet and the block converts almost all of it to heat.
 
  • #11
~christina~
Gold Member
717
0
Is the way I did part c correct?

c) max acceleration of system
F= ma
F= -kx

(m+ M)a= kx

so


[tex]a= \frac{0.42214x0.075m} {(0.00972kg + 10.00kg)} = 0.0031629m/s^2 [/tex]
 
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  • #12
~christina~
Gold Member
717
0
Can someone help me and see if the last part is correct (post above this one)?

I just need to see if that is correct.

THANK YOU
 
  • #13
kamerling
454
0
Is the way I did part c correct?

c) max acceleration of system
F= ma
F= -kx

(m+ M)a= kx

so


[tex]a= \frac{0.42214x0.075m} {(0.00972kg + 10.00kg)} = 0.0031629m/s^2 [/tex]
you need to substitute the new value of k.
 
  • #14
~christina~
Gold Member
717
0
you need to substitute the new value of k.

Oh I did that on my paper.


Thank you. (I just wanted to know if the method of solving that was right)
 

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