In a supermarket,you place a 22.3-N (around 5 lb)bag of oranges on a scale,and the scale starts to oscillate at 2.7 Hz.What is the force constant (spring constant)of the spring of the scale?
A)650 N/m
B)600 N/m
C)330 N/m
D)820 N/m
E)410 N/m

Question

Quizplus · Accepted Answer

The force constant (spring constant) $k$ can be found using the formula for the period $T$ of a mass-spring system: $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. However, since we are given the frequency $f = 2.7 \, \text{Hz}$, we can use the relation $f = \frac{1}{T}$ to find $T$, and then solve for $k$. The mass $m$ can be found by dividing the weight of the bag by the acceleration due to gravity ($g = 9.8 \, \text{m/s}^2$). The weight of the bag is $22.3 \, \text{N}$, so $m = \frac{22.3 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 2.27 \, \text{kg}$.Given $f = 2.7 \, \text{Hz}$, we have $T = \frac{1}{f} = \frac{1}{2.7} \approx 0.37 \, \text{s}$.Substituting $T$ back into the formula for $T$ in terms of $m$ and $k$, and solving for $k$, we get:$$0.37 = 2\pi\sqrt{\frac{2.27}{k}}$$Solving for $k$, we find $k \approx 650 \, \text{N/m}$.

In a Supermarket,you Place a 22