A mass m at the end of a spring oscillates with a frequency of 0.80 Hz . When an additional 690 g mass is added to m, the frequency is 0.55 Hz .
What is the value of m?
Express your answer to two significant figures and include the appropriate units.
The Correct Answer and Explanation is :
We are tasked with finding the value of the original mass mm based on the frequencies before and after an additional mass is added. To solve this, we will use the formula for the frequency of oscillation of a mass-spring system: f=12πkmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}
Where:
- ff is the frequency of oscillation,
- kk is the spring constant, and
- mm is the mass.
Step 1: Relating the two frequencies
Before the additional mass is added, the frequency is f1=0.80 Hzf_1 = 0.80 \, \text{Hz}, and after adding the mass m2=690 g=0.690 kgm_2 = 690 \, \text{g} = 0.690 \, \text{kg}, the frequency becomes f2=0.55 Hzf_2 = 0.55 \, \text{Hz}.
We can use the formula for frequency to relate the spring constant kk and the masses involved: f1=12πkm(1)f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \quad \text{(1)} f2=12πkm+m2(2)f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m + m_2}} \quad \text{(2)}
We can eliminate kk by dividing equation (2) by equation (1): f2f1=mm+m2\frac{f_2}{f_1} = \sqrt{\frac{m}{m + m_2}}
Squaring both sides: (f2f1)2=mm+m2\left( \frac{f_2}{f_1} \right)^2 = \frac{m}{m + m_2}
Substitute the values of f1f_1, f2f_2, and m2m_2: (0.550.80)2=mm+0.690\left( \frac{0.55}{0.80} \right)^2 = \frac{m}{m + 0.690} (0.6875)2=mm+0.690\left( 0.6875 \right)^2 = \frac{m}{m + 0.690} 0.472=mm+0.6900.472 = \frac{m}{m + 0.690}
Step 2: Solve for mm
Now solve for mm: 0.472(m+0.690)=m0.472 (m + 0.690) = m 0.472m+0.32568=m0.472m + 0.32568 = m 0.528m=0.325680.528m = 0.32568 m=0.325680.528m = \frac{0.32568}{0.528} m≈0.616 kgm \approx 0.616 \, \text{kg}
Final Answer:
The value of the original mass is approximately 0.62 kg (to two significant figures).
Explanation:
The system oscillates with a frequency determined by the spring constant kk and the mass attached to the spring. By comparing the frequencies before and after adding a mass, we can derive a relationship between the initial mass and the added mass. By solving the equation, we find the original mass.