Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
5. The two springs from the problem above are reconnected to mass m as shown. Now what is the frequency of oscillation? 2 ellue m Fee ull

5. The two springs from the problem above are reconnected to mass m as shown. Now what is the frequency of oscillation? 2 ellue m Fee ull

icon
Related questions
Question
100%
Can someone please explain it to me ASAP?!!! This is periodic motion
**Problem: Determining the Frequency of Oscillation** **Question:** The two springs from the problem above are reconnected to mass \( m \) as shown. Now, what is the frequency of oscillation? **Diagram:** The diagram shows a mass \( m \) connected between two springs with spring constants \( k_1 \) and \( k_2 \). The spring \( k_1 \) is attached to the left of the mass, and the spring \( k_2 \) is attached to the right of the mass. Both springs are fixed at the opposite ends, forming a horizontal system. To determine the frequency of oscillation (\( \omega \)) of the mass \( m \), we can start by noting that the springs are in series. The effective spring constant \( k \) of two springs in series is given by: \[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \] Simplifying this, we get: \[ k = \frac{k_1 k_2}{k_1 + k_2} \] The frequency of oscillation is given by the formula: \[ \omega = \sqrt{\frac{k}{m}} \] Substituting the effective spring constant \( k \) into the equation, we obtain the frequency of oscillation: \[ \omega = \sqrt{\frac{\frac{k_1 k_2}{k_1 + k_2}}{m}} = \sqrt{\frac{k_1 k_2}{m(k_1 + k_2)}} \] Thus, the frequency of oscillation for the mass \( m \) connected to two springs with spring constants \( k_1 \) and \( k_2 \) is: \[ \boxed{\omega = \sqrt{\frac{k_1 k_2}{m(k_1 + k_2)}}} \] This formula allows us to find the natural frequency of the system based on the given spring constants and mass.
Transcribed Image Text:**Problem: Determining the Frequency of Oscillation** **Question:** The two springs from the problem above are reconnected to mass \( m \) as shown. Now, what is the frequency of oscillation? **Diagram:** The diagram shows a mass \( m \) connected between two springs with spring constants \( k_1 \) and \( k_2 \). The spring \( k_1 \) is attached to the left of the mass, and the spring \( k_2 \) is attached to the right of the mass. Both springs are fixed at the opposite ends, forming a horizontal system. To determine the frequency of oscillation (\( \omega \)) of the mass \( m \), we can start by noting that the springs are in series. The effective spring constant \( k \) of two springs in series is given by: \[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \] Simplifying this, we get: \[ k = \frac{k_1 k_2}{k_1 + k_2} \] The frequency of oscillation is given by the formula: \[ \omega = \sqrt{\frac{k}{m}} \] Substituting the effective spring constant \( k \) into the equation, we obtain the frequency of oscillation: \[ \omega = \sqrt{\frac{\frac{k_1 k_2}{k_1 + k_2}}{m}} = \sqrt{\frac{k_1 k_2}{m(k_1 + k_2)}} \] Thus, the frequency of oscillation for the mass \( m \) connected to two springs with spring constants \( k_1 \) and \( k_2 \) is: \[ \boxed{\omega = \sqrt{\frac{k_1 k_2}{m(k_1 + k_2)}}} \] This formula allows us to find the natural frequency of the system based on the given spring constants and mass.
Expert Solution
steps

Step by step

Solved in 4 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer