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A spring has an unstretched length of 20 cm. A 100g mass hanging from the spring stretches it to an equilibrium length of 30 cm. For the data given above, what is the frequency of the oscillation?

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Final answer:

The frequency of the oscillation for a mass hanging from a spring can be found by determining the spring constant using Hooke's law and the weight of the mass, then calculating the angular frequency, and finally finding the frequency by dividing the angular frequency by 2π.

Step-by-step explanation:

To determine the frequency of the oscillation for a mass hanging from a spring, we first need to calculate the spring constant (k) using Hooke's law, F = kx, where F represents the force applied to the spring, and x is the displacement from the equilibrium position. The force exerted by the hanging mass (weight) can be calculated using the equation F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s2). In this case, F = 0.1 kg * 9.8 m/s2 = 0.98 N, and the displacement x is 10 cm or 0.1 m since the spring stretches from 20 cm to 30 cm. Therefore, k = F/x = 0.98 N / 0.1 m = 9.8 N/m.

Once we have the spring constant, we can calculate the angular frequency (ω) using the formula ω = √(k/m), and the frequency (f) is given by f = ω/(2π). Substituting the values, we get ω = √(9.8 N/m / 0.1 kg) and f = ω/(2π), resulting in a frequency f for the oscillating system.

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