Question

an object with mass ( m ) attached to a horizontal spring with spring constant ( k ) undergoes simple harmonic motion with amplitude ( a ) and period ( t ). which of the following expressions is equal to the maximum speed of the block?
a ( \frac{4a}{t} )
b ( sqrt{\frac{k}{m}} a )
c ( \frac{k}{m} a^2 )
d ( \frac{1}{2} ka^2 )

Explanation:

Step1: Recall energy conservation in SHM

In simple harmonic motion (SHM) of a spring - mass system, the total mechanical energy is conserved. The potential energy of the spring is given by \(U=\frac{1}{2}kx^{2}\) and the kinetic energy of the mass is given by \(K = \frac{1}{2}mv^{2}\). At the equilibrium position (where the speed is maximum), the potential energy is zero and the kinetic energy is maximum. At the extreme position (where \(x = A\), the amplitude), the kinetic energy is zero and the potential energy is maximum. So, by conservation of energy, \(\frac{1}{2}kA^{2}=\frac{1}{2}mv_{max}^{2}\).

Step2: Solve for \(v_{max}\)

Start with the energy - conservation equation \(\frac{1}{2}kA^{2}=\frac{1}{2}mv_{max}^{2}\).
First, we can cancel out the \(\frac{1}{2}\) from both sides of the equation: \(kA^{2}=mv_{max}^{2}\).
Then, divide both sides by \(m\): \(\frac{kA^{2}}{m}=v_{max}^{2}\).
Take the square root of both sides: \(v_{max}=\sqrt{\frac{k}{m}}A\).

We can also analyze option A. The formula for the speed in SHM is \(v = \omega\sqrt{A^{2}-x^{2}}\), and \(\omega=\frac{2\pi}{T}\). The maximum speed \(v_{max}=\omega A=\frac{2\pi A}{T}\approx\frac{6.28A}{T}\), which is not equal to \(\frac{4A}{T}\), so option A is incorrect.

Option C has units of \(\frac{N/m}{kg}\times m^{2}=\frac{(kg\cdot m/s^{2})/m}{kg}\times m^{2}=m^{2}/s^{2}\), and the square - root of that would be \(m/s\) (the unit of speed), but the expression \(\frac{k}{m}A^{2}\) is not equal to the maximum speed (we saw from energy conservation that the correct expression comes from \(\frac{1}{2}kA^{2}=\frac{1}{2}mv_{max}^{2}\), not from this form).

Option D, \(\frac{1}{2}kA^{2}\), has units of energy (joules), not speed (m/s), so it is incorrect.

Answer:

B. \(\boldsymbol{\sqrt{\frac{k}{m}}A}\)