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olga nikolaevna [1]

3 years ago

9

An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

otion. What is the magnitude of the acceleration of the object when it is at its maximum displacement of 0.10 m?

1 answer:

Sergeeva-Olga [200]3 years ago

5 0

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement 0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

= \frac{8 *0 .10}{0.40}

a = 2 m/s2

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Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

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Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

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Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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