Questions: A spring stretches by 0.0154 m when a 4.91-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f=5.52 Hz?

Solution Steps

First, we need to find the spring constant \( k \) using Hooke's Law, which states that the force \( F \) exerted by a spring is proportional to the displacement \( x \) from its equilibrium position:

\[ F = kx \]

The force exerted by the spring is equal to the weight of the object, which is \( mg \), where \( m = 4.91 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). The displacement \( x \) is given as \( 0.0154 \, \text{m} \).

\[ 4.91 \times 9.81 = k \times 0.0154 \]

Solving for \( k \):

\[ k = \frac{4.91 \times 9.81}{0.0154} = 3127.4026 \, \text{N/m} \]

The frequency \( f \) of a mass-spring system is given by:

\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

We need to find the mass \( m \) that results in a frequency of \( 5.52 \, \text{Hz} \).

Rearrange the frequency formula to solve for \( m \):

\[ m = \frac{k}{(2\pi f)^2} \]

Substitute the known values:

\[ m = \frac{3127.4026}{(2\pi \times 5.52)^2} \]

Calculate \( m \):

\[ m = \frac{3127.4026}{(34.6656)^2} = \frac{3127.4026}{1201.788} = 2.6021 \, \text{kg} \]

Final Answer