Question

current attempt in progress

objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. the spring constant of spring 1 is 155 n/m. the motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. the magnitude of the maximum velocity is the same in each case. find the spring constant of spring 2.

k=

save for later

using multiple attempts will impact your score.

20% score reduction after attempt 2

attempts: 0 of 3 used

submit answer

Explanation:

Step1: Write the formula for maximum velocity

The maximum velocity in simple - harmonic motion is \(v_{max}=\omega A\), where \(\omega=\sqrt{\frac{k}{m}}\). So \(v_{max}=\sqrt{\frac{k}{m}}A\)

Step2: Set up the equation for the two springs

Let \(k_1 = 155\space N/m\), \(A_1 = 2A_2\), \(v_{max1}=v_{max2}\).
Since \(v_{max}=\sqrt{\frac{k}{m}}A\) and \(v_{max1} = v_{max2}\), we have \(\sqrt{\frac{k_1}{m}}A_1=\sqrt{\frac{k_2}{m}}A_2\)

Step3: Substitute \(A_1 = 2A_2\) into the equation

\(\sqrt{\frac{k_1}{m}}\times(2A_2)=\sqrt{\frac{k_2}{m}}A_2\). Cancel out \(A_2\) and \(\sqrt{m}\) (since \(m\) is non - zero) from both sides of the equation. We get \(2\sqrt{k_1}=\sqrt{k_2}\)

Step4: Square both sides of the equation

\((2\sqrt{k_1})^2 = k_2\). Using the formula \((ab)^2=a^{2}b^{2}\), we have \(4k_1=k_2\)

Step5: Calculate \(k_2\)

Substitute \(k_1 = 155\space N/m\) into \(k_2 = 4k_1\). So \(k_2=4\times155 = 620\space N/m\)

Answer:

\(620\space N/m\)