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sleet_krkn [62]
3 years ago
14

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

splaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?
Physics
1 answer:
AleksandrR [38]3 years ago
4 0

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

T = 2\pi \sqrt{\frac{m}{k}}

Where,

m = Mass

k = Spring constant

T = 2\pi \sqrt{\frac{0.35}{28.58}}

T = 0.685s

Therefore the period of the oscillations is 0.685s

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10
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Time interval;Δt ≈ 37 seconds

Explanation:

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Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
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0.853 m/s

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make v the subject of the equation,

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