Block- spring- bullet- harmonic motion

In summary: M)} {4\pi^2} So, it should bea= \frac{0.775Hz^2(0.00972kg + 10.00kg)} {(4\pi^2)} a= 0.0031629 m/s^2 Is that correct?Also, does anyone know what the units for k would be?Thanks.In summary, the assignment is to measure the muzzle velocity of a bullet using a 10kg block attached to a Hooke's Law spring. Preliminary data shows a 9.72g bullet colliding with the block and compressing the spring by 7.50cm
  • #1

~christina~

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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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  • #2
~christina~ said:
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
[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
kamerling said:
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?

Doc Al said:
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
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
~christina~ said:
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
~christina~ said:
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
kamerling said:
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
kamerling said:
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
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
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
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
~christina~ said:
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
kamerling said:
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)
 

1. What is harmonic motion?

Harmonic motion is a type of periodic motion in which an object moves back and forth around a central equilibrium point due to the forces of a spring or other elastic material.

2. How does a block-spring-bullet system exhibit harmonic motion?

In a block-spring-bullet system, the bullet is shot into the block, causing it to compress the spring. The spring then exerts a force on the block, causing it to oscillate back and forth in a harmonic motion.

3. What factors affect the frequency of harmonic motion in a block-spring-bullet system?

The frequency of harmonic motion in a block-spring-bullet system is affected by the mass of the block and bullet, the spring constant of the spring, and the initial velocity of the bullet.

4. How is the amplitude of harmonic motion in a block-spring-bullet system related to the initial velocity of the bullet?

The amplitude of harmonic motion in a block-spring-bullet system is directly proportional to the initial velocity of the bullet. This means that a higher initial velocity will result in a larger amplitude of motion.

5. What is the equation for calculating the period of harmonic motion in a block-spring-bullet system?

The period of harmonic motion in a block-spring-bullet system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the block and bullet, and k is the spring constant of the spring.

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