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FromTheMoon [43]
3 years ago
10

Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a sta

tic method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.

Physics
1 answer:
yuradex [85]3 years ago
7 0

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

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Answer : The correct option is (d) 2.73 m

Explanation :

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2

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Now put all the given values in the above equation, we get:

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The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
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Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

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           f = w / 2π

           f = 3 / 2π

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frequency and period are related

           T = 1 / f

           T = 1 / 0.477

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e) the phase constant

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          v = dx / dt

         

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h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

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for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

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