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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 160 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.
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Explanation:

Step1: Recall the formula for maximum velocity in simple harmonic motion

The formula for the maximum velocity \(v_{max}\) in simple harmonic motion is \(v_{max}=\omega A\), where \(\omega = \sqrt{\frac{k}{m}}\) (from \(F=-kx = ma\), \(a =-\frac{k}{m}x\), and for SHM \(a =-\omega^{2}x\)) and \(A\) is the amplitude. So \(v_{max}=A\sqrt{\frac{k}{m}}\)

Step2: Set up equations for both springs

Let \(k_1 = 160\space N/m\), \(A_1 = 2A_2\), \(v_{max1}=v_{max2}\), and \(m_1 = m_2=m\) (since masses are equal).
For spring 1: \(v_{max1}=A_1\sqrt{\frac{k_1}{m}}\)
For spring 2: \(v_{max2}=A_2\sqrt{\frac{k_2}{m}}\)
Since \(v_{max1} = v_{max2}\), we have \(A_1\sqrt{\frac{k_1}{m}}=A_2\sqrt{\frac{k_2}{m}}\)

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

Substituting \(A_1 = 2A_2\) into \(A_1\sqrt{\frac{k_1}{m}}=A_2\sqrt{\frac{k_2}{m}}\), the \(m\) cancels out.
\(2A_2\sqrt{\frac{k_1}{m}}=A_2\sqrt{\frac{k_2}{m}}\)
Divide both sides by \(A_2\) (since \(A_2
eq0\)), we get \(2\sqrt{k_1}=\sqrt{k_2}\)

Step4: Solve for \(k_2\)

Square both sides of the equation \(2\sqrt{k_1}=\sqrt{k_2}\)
\((2\sqrt{k_1})^2=k_2\)
\(4k_1 = k_2\)
Substitute \(k_1 = 160\space N/m\)
\(k_2=4\times160\space N/m\)

Answer:

\(640\space N/m\)