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dusya [7]
4 years ago
13

A mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. If the mass of the object is 0.20 kg and the sp

ring constant is 130 N/m, what is the frequency?
Physics
1 answer:
Dmitry [639]4 years ago
5 0

Answer:

Frequency, f = 3.73Hz

Explanation:

The frequency of a simple harmonic 6is given by:

f = w/2pi

But w= Sqrt( k/m)

Where k is the spring constant

And m is the mass

Given:

Mass=0.20kg

Spring constant, k=130N/m

w= Sqrt(130/0.20)

w= Sqrt(650)

w= 25.50 m

Frequency, f = w/2pi

f = 25.50/(2×3.142)

f = 3.73Hz

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Answer:

(a) the mechanical energy of the system, U = 0.1078 J

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Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

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(b) the maximum speed of the object

V_{max} =\omega*A=  \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s

(c) the maximum acceleration of the object

a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2

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