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bogdanovich [222]
3 years ago
13

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

orce acting on it? (b) If the oscillations are produced by a spring, what is the springconstant?
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

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Option E is correct 310N

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Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

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