A mass attached to a simple horizontal spring undergoes SHM. What is its period T in terms of mass m and spring constant k?

A mass attached to a simple horizontal spring undergoes SHM. What is its period T in terms of mass m and spring constant k?

• A. T = 2π sqrt(m/k)
• B. T = 2π sqrt(k/m)
• C. T = π sqrt(m/k)
• D. T = sqrt(mk)/π

Answer: (A)

When a mass on a horizontal spring undergoes simple harmonic motion, the motion is governed by Hooke’s law F = -kx and Newton’s second law m a = F. This leads to the equation of motion m x'' = -k x, or x'' + (k/m) x = 0. That form tells us the angular frequency is ω = sqrt(k/m). The period, which is the time for one full oscillation, is T = 2π/ω = 2π sqrt(m/k). So the period grows with mass and decreases with spring stiffness, as expected: a heavier mass takes longer to complete a cycle, while a stiffer spring makes the cycle faster. The other expressions don’t fit the period: sqrt(k/m) would be the angular frequency, not the period; missing a factor of 2 gives an incorrect time scale; and sqrt(mk)/π mixes the parameters in a way that does not match the correct dimensional and physical dependence.

When a mass on a horizontal spring undergoes simple harmonic motion, the motion is governed by Hooke’s law F = -kx and Newton’s second law m a = F. This leads to the equation of motion m x'' = -k x, or x'' + (k/m) x = 0. That form tells us the angular frequency is ω = sqrt(k/m). The period, which is the time for one full oscillation, is T = 2π/ω = 2π sqrt(m/k). So the period grows with mass and decreases with spring stiffness, as expected: a heavier mass takes longer to complete a cycle, while a stiffer spring makes the cycle faster.

The other expressions don’t fit the period: sqrt(k/m) would be the angular frequency, not the period; missing a factor of 2 gives an incorrect time scale; and sqrt(mk)/π mixes the parameters in a way that does not match the correct dimensional and physical dependence.