# M3 questionWatch

This discussion is closed.
#1
Exercise 3E, q7. I get the extension to be 0.4+0.02g, but the time period as 0.222 not 0.295

question is :
Particle P of mass 0.3kg is attached one end of a light elastic string of modulus 9N. P is pulled down a further 0.4 m and released. Calculate the time that elapses before the string becomes slack and the further time before the string is once more taught.
0
14 years ago
#2
What is the natural length of the string?
0
14 years ago
#3
(Original post by Jonny W)
What is the natural length of the string?
can you not work it out using stress/strain/ etc? personally i aint finished m2 yet so i could be talking turd...
0
#4
(Original post by Jonny W)
What is the natural length of the string?
sorry, l=0.6m
0
14 years ago
#5
Extension of string in equilibrium
= 0.3gl/9
= 0.02g . . . . . as you said

Downwards force on P when P is a distance x below its equilibrium position
= 0.3g - 9(x + 0.02g)/l
= -15x

0.3 x'' = -15x
x'' = -50x . . . . . SHM equation
x(t) = A cos(sqrt(50)t + k) . . . . . standard solution
x(t) = 0.4 cos(sqrt(50)t + k) . . . . . since amplitude of solution is 0.4
x(t) = 0.4 cos(sqrt(50)t) . . . . . since x(0) = 0.4

The string becomes slack for the first time when x = -0.02g. At that moment,

0.4 cos(sqrt(50)t) = -0.02g

t
= arccos(-0.02g/0.4)/sqrt(50)
= 0.295

--

The second part is just a projectile question (the string is slack, so it doesn't affect the motion).
0
#6
(Original post by Jonny W)
Extension of string in equilibrium
= 0.3gl/9
= 0.02g . . . . . as you said

Downwards force on P when P is a distance x below its equilibrium position
= 0.3g - 9(x + 0.02g)/l
= -15x

0.3 x'' = -15x
x'' = -50x . . . . . SHM equation
x(t) = A cos(sqrt(50)t + k) . . . . . standard solution
x(t) = 0.4 cos(sqrt(50)t + k) . . . . . since amplitude of solution is 0.4
x(t) = 0.4 cos(sqrt(50)t) . . . . . since x(0) = 0.4

The string becomes slack for the first time when x = -0.02g. At that moment,

0.4 cos(sqrt(50)t) = -0.02g

t
= arccos(-0.02g/0.4)/sqrt(50)
= 0.295

--

The second part is just a projectile question (the string is slack, so it doesn't affect the motion).
hmm yes. thank you. I keep forgetting that the equilibrium point is 0.02g down.
0
#7
The second part of the question is this:
Calculate the further time before the string becomes taught.

Pls help :-)
0
14 years ago
#8
(Original post by lgs98jonee)
The second part of the question is this:
Calculate the further time before the string becomes taught.

Pls help :-)
The further time before the string becomes taut is the time for it to travel 2.(natural length) when moving as a projectile.
0
#9
(Original post by Gaz031)
The further time before the string becomes taut is the time for it to travel 2.(natural length) when moving as a projectile.
i know :-)/ but i can not get the speed right. And does it not travel 2l+the extension due to gravity?
0
14 years ago
#10
x(t) = 0.4 cos(sqrt(50)t)
x'(t) = -0.4sqrt(50) sin(sqrt(50)t)

Velocity of P when the string becomes slack
= x'(0.2946)
= -2.4656 m/s

Time for string to become taut again
= 2*2.4656/9.8
= 0.503 s
0
#11
Thank you very much. As you can see, I am fairly hopeless at M3. So i have another couple of questions.
How do I find the time period of the oscillation of a jack in the box, when I only know:
the mass
the amount the string is compressed by due to gravity
the amount it is pushed down for the oscillation to start
0
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