A mass m at the end of a spring oscillates with a frequency of 0.78 Hz. When an additional 770 g mass is added to m, the frequency is 0.61 Hz. What is the value of m?
A mass m at the end of a spring oscillates with a frequency of 0.78 Hz. When an additional 770 g mass is added to m, the frequency is 0.61 Hz.
What is the value of m?
The Correct Answer and Explanation is :
To solve this problem, we need to use the formula for the frequency of a mass-spring system:
[
f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}
]
Where:
- ( f ) is the frequency,
- ( k ) is the spring constant,
- ( m ) is the mass attached to the spring.
Step 1: Express the frequencies for both cases
We are given two scenarios:
- When the mass is ( m ), the frequency is ( f_1 = 0.78 ) Hz.
- When an additional 770 g (0.770 kg) mass is added, the total mass becomes ( m + 0.770 ) kg, and the frequency is ( f_2 = 0.61 ) Hz.
We can write the frequency equations for both cases:
[
f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \quad \text{(for mass ( m ))}
]
[
f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m + 0.770}} \quad \text{(for mass ( m + 0.770 ))}
]
Step 2: Relating the frequencies
Dividing the second equation by the first gives:
[
\frac{f_2}{f_1} = \frac{\sqrt{\frac{k}{m + 0.770}}}{\sqrt{\frac{k}{m}}}
]
Simplifying:
[
\frac{f_2}{f_1} = \sqrt{\frac{m}{m + 0.770}}
]
Substitute the known values for ( f_1 ) and ( f_2 ):
[
\frac{0.61}{0.78} = \sqrt{\frac{m}{m + 0.770}}
]
Squaring both sides:
[
\left(\frac{0.61}{0.78}\right)^2 = \frac{m}{m + 0.770}
]
Now calculate the left-hand side:
[
\left(\frac{0.61}{0.78}\right)^2 \approx 0.6025
]
So the equation becomes:
[
0.6025 = \frac{m}{m + 0.770}
]
Step 3: Solve for ( m )
Cross-multiply to solve for ( m ):
[
0.6025(m + 0.770) = m
]
Distribute the 0.6025:
[
0.6025m + 0.463925 = m
]
Now move the terms with ( m ) to one side:
[
0.463925 = m – 0.6025m
]
Simplifying:
[
0.463925 = 0.3975m
]
Solve for ( m ):
[
m = \frac{0.463925}{0.3975} \approx 1.167 \, \text{kg}
]
Final Answer:
The value of the mass ( m ) is approximately 1.17 kg.
Explanation:
The oscillation frequency of a mass-spring system depends on the mass and the spring constant. By using the relationship between frequency and mass, we set up a ratio of the frequencies before and after adding the extra mass. After solving the equation, we determined the original mass of the object. This type of problem helps demonstrate how frequency changes with mass in harmonic motion.