When is the magnitude of the acceleration of a mass on a spring at its maximum value?

Physics

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

Explanation:

Now we know that Force is the rate of change of momentum meaning

F= mv/t

But

mv/t = Ke

v/t = ke/m

a= ke/m

Where a is acceleration

K is constant of proportionality of tension on a spring

e is the extension substended by a string.

From the formula of acceleration we can see that as mass decreases acceleration increases so we can see that m = 1

We would have a maximum value of acceleration.

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Snowboarder Jump—Energy and Momentum

soldier1979 [14.2K]

Answer:

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Explanation:

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3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$