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Difficulty: MediumSimple Harmonic Motion

A 0.40 kg0.40\text{ kg} mass attached to a light helical spring undergoes simple harmonic motion on a smooth horizontal surface. If the force constant of the spring is 160 N/m160\text{ N/m} and the amplitude of oscillation is 0.05 m0.05\text{ m}, what is the maximum speed of the mass?

  1. 1.0 m/s1.0\text{ m/s}Answer
  2. B
    20.0 m/s20.0\text{ m/s}
  3. C
    0.40 m/s0.40\text{ m/s}
  4. D
    4.00 m/s4.00\text{ m/s}

Answer

The maximum speed of the mass is 1.0 m/s1.0\text{ m/s}.
The angular frequency of the mass-spring system is calculated using \(\omega = \sqrt{k/m} = \sqrt{160/0.40} = 20\text{ rad/s}\). Multiplying this by the amplitude \(A = 0.05\text{ m}\) yields a maximum speed of \(v_{\text{max}} = 1.0\text{ m/s}\).

Step-by-Step Solution

1
Calculate the angular frequency (\(\omega\)) of the mass-spring system.
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160\text{ N/m}}{0.40\text{ kg}}} = \sqrt{400} = 20\text{ rad/s}\)
The angular frequency of a spring-mass oscillator depends on the spring constant and the mass.
2
Determine the maximum speed (\(v_{\text{max}}\)) using the amplitude.
\(v_{\text{max}} = \omega A = 20\text{ rad/s} \times 0.05\text{ m} = 1.0\text{ m/s}\)
In simple harmonic motion, maximum speed occurs at the equilibrium position and equals the product of angular frequency and amplitude.

Key Concept

Maximum velocity in simple harmonic motion for a mass-spring system
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Topics

Simple Harmonic Motion

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