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3 years ago

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A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

cy is 0.63 hz. what is the value of m?

1 answer:

MrMuchimi3 years ago

5 0

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²

Then we know that k is constant for both trials, we have:
k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

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kari74 [83]

Answer:

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Explanation:

Given:

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a.

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$w=m.g$

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$w=9.8\times10^{-4}\ N$

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b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

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$F=9.8\times10^{-4}\ N$

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<u>Answer</u>

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