Question

Two packages are placed on a spring scale whose plate weighs \(10 \mathrm{lb}\) and whose stiffness is \(50 \mathrm{lb} / \mathrm{in}\). When one package is accidentally knocked off the scale the remaining package is observed to oscillate through 3 cycles per second. What is the weight of the remaining package?

Short answer

Answer: The approximate weight of the remaining package is 5.66 lb.

Step-by-step solution

Determine the combined weight of the packages on the scale

First, let W1 and W2 be the weight of the first and second package, respectively. The combined weight of both packages is given by W1 + W2. The spring with both packages will compress by an amount x, such that W1 + W2 = kx, where k is the stiffness of the spring (50 lb/in).

Determine the combined mass of the packages with the given weight

We know that weight = mass × gravity. Thus, the mass M1 and M2 of the first and second package can be calculated as M1 = W1/32.2 ft/s² and M2 = W2/32.2 ft/s² (Assuming standard gravity is ~32.2 ft/s²).

Obtain the total mass on the spring

When both packages are on the spring, the total mass on the spring is M = M1 + M2.

Calculate the spring's natural frequency

We are given that the natural frequency f of the remaining package after knocking off one of the packages is 3 cycles per second. Using the formula for the frequency of oscillation under spring mass system, f = (1 / (2 * pi)) * sqrt(k/M), where k is the spring stiffness (50 lb/in) and M is the mass of the remaining package.

Solve for the weight of the remaining package, W2

First, we need to rearrange the formula for frequency in terms of M: $$ M = \frac{k}{{(2\cdot\pi\cdot f)}^2} $$From the previous steps we know W1+W2= kx and the mass of remaining package is M2=W2/32.2 = $$ \frac{k}{{(2\cdot\pi\cdot f)}^2} $$. Now, multiplying both sides by 32.2, we obtain: $$ W2 = 32.2\cdot\frac{k}{{(2\cdot\pi\cdot f)}^2} $$Substitute the given values for k, f, and 32.2 ft/s² for gravity, and solve for W2.

Calculate the weight of the remaining package

Finally, by substituting the given values (k=50 lb/in and f=3 cycles/s) into the equation, we get: $$ W2 = 32.2\cdot\frac{50}{(2\cdot(3.1416)\cdot 3)^2} $$After evaluating the expression, you get W2 to be approximately 5.66 lb. Therefore, the weight of the remaining package is approximately 5.66 lb.