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By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s?

a) ±0.01 kg, 4%
b) ±0.02 kg, 8%
c) ±0.03 kg, 12%
d) ±0.04 kg, 16%

User CamiloEr
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1 Answer

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Final answer:

The leeway in the selection of the mass of the object would be ±0.01 kg or ±2%.

Step-by-step explanation:

To calculate the leeway in the selection of the mass of the object, we need to find the range of masses that will result in a new period between 1.99 s and 2.01 s. Since the equation for the period of a simple harmonic oscillator is T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant, we can rearrange the equation to solve for m. From the given information, we know that the original period (T1) is 1.50 s and the desired new period (T2) is between 1.99 s and 2.01 s.

Using the equation T1 = 2π√(m1/k) and T2 = 2π√(m2/k), we can substitute the values and solve for m2 and m1 respectively. Then we can find the difference in mass (Δm = m2 - m1) in order to determine the leeway in mass.

After performing the calculations, we find that the leeway in mass is ±0.01 kg or ±2%.

User Riwall
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