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Talja [164]
4 years ago
10

Suppose a spring has a relaxed length of 25.1 cm. The simulation refers to this as the natural length. This is the length of the

spring when it is not stretched nor compressed by any force. When you hang a mass of 250 g on the vertical spring, the spring stretches to a new length of 33.9 cm. What is an estimate of the spring constant, k, of the spring? (Enter your answer in N/m.) Measuring only one value is not a good way to get the value, but you can an idea of the approximate size.
Physics
1 answer:
Leviafan [203]4 years ago
7 0
The anwser is 30096 hope it helps bud
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3 years ago
Read 2 more answers
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Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

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grandymaker [24]
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