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Interactive physics simulator

Velocity in SHM

Investigate the speed and direction of simple harmonic oscillators. Toggle between real-time Phase Space ellipse plotting, perpendicular phasor circular projections, and dynamic Kinetic vs. Potential Energy balances.

Velocity in SHM Laboratory ⓘ

Modify physical parameters in the right panel. Watch the live velocity (green), displacement (blue), and force (red) vectors synchronize on stage.

Oscillating

Live Velocity Telemetry

Elapsed Time (t)
0.00 s
Displacement (x)
0.00 m
Velocity (v)
0.00 m/s
Kinetic Energy (K)
0.00 J

Understanding Velocity in Simple Harmonic Motion

In physics, **velocity** represents the rate of change of position of an oscillating object with respect to time. Unlike displacement, which measures how far the object is from equilibrium, velocity describes how fast and in what direction it is moving at any given instant.

When a body undergoes simple harmonic motion, its velocity changes continuously. Mathematically, by differentiating the displacement function x(t) = A cos(ωt + φ), we obtain the velocity function: v(t) = -Aω sin(ωt + φ). This function shows that velocity varies sinusoidally, with a maximum value of vmax = Aω. Because velocity is written in terms of a sine function while displacement uses a cosine function, velocity is exactly π/2 radians (90°) out of phase with displacement, meaning it leads displacement by a quarter-cycle.

Key Principles

Core characteristics of SHM velocity:

  • Equilibrium Maximum: Velocity is at its peak magnitude (v = ±vmax) when crossing the equilibrium center (x = 0).
  • Extreme Turning Points: Velocity drops to zero (v = 0) at the maximum displacements (x = ±A), where the oscillator pauses to reverse.
  • Phase Shift Lead: The velocity vector leads the displacement vector by π/2 radians (90°), shifting the speed peaks a quarter cycle ahead.
  • Energy Conservation: Kinetic energy is maximum when speed is maximum, perfectly matching the minimum potential energy at x = 0.

Formulas & Relations

Mathematical definitions of velocity:

  • Time-Dependent Velocity: v(t) = -Aω sin(ωt + φ). The derivative of displacement.
  • Position-Dependent Velocity: v(x) = ±ω√(A² - x²). Describes velocity at any displacement point x.
  • Maximum Velocity: vmax = Aω. Peak velocity occurring at x = 0.
  • Angular Frequency: ω = 2πf = √(k/m). Determines the speed of rotation/cycles.
  • Kinetic Energy: K = {1/2} m v². Represents the moving energy of the mass bob.

Solved Examples

A particle oscillates in SHM with a displacement described by: x(t) = 0.40 cos(4.0π t + π/6), where x is in meters and t is in seconds. (a) Derive the equation for velocity as a function of time. (b) Find the instantaneous velocity of the particle at t = 0.50 seconds.
  1. Recall that velocity is the first derivative of displacement with respect to time: v(t) = dx/dt.
  2. Differentiate x(t) = 0.40 cos(4.0π t + π/6). Note that d/dt [cos(u)] = -sin(u) · du/dt.
  3. Calculate: v(t) = -0.40 · (4.0π) · sin(4.0π t + π/6) = -1.6π sin(4.0π t + π/6) m/s.
  4. This is the velocity equation. The maximum speed is v_max = 1.6π ≈ 5.03 m/s.
  5. Part (b): Substitute t = 0.50 s into the velocity equation.
  6. v(0.50) = -1.6π sin(4.0π · 0.50 + π/6) = -1.6π sin(2.0π + π/6).
  7. Using the periodicity of sine: sin(2.0π + π/6) = sin(π/6) = 0.50.
  8. Multiply parameters: v(0.50) = -1.6π · 0.50 = -0.8π ≈ -2.51 m/s.
  9. The velocity at t = 0.50 s is approximately -2.51 m/s (moving in the negative direction).

Answer: v(t) = -1.6π sin(4.0π t + π/6) m/s, v(0.50) ≈ -2.51 m/s

A mass-spring system consists of a 0.50 kg mass attached to a spring with k = 200 N/m. The system is stretched by 10.0 cm from its equilibrium position and released from rest at t = 0. (a) Calculate the maximum velocity of the oscillator. (b) Calculate the magnitude of the velocity when the mass is at displacement x = 6.0 cm.
  1. Identify the parameters: mass m = 0.50 kg, spring constant k = 200 N/m, amplitude A = 10.0 cm = 0.10 m.
  2. Calculate the angular frequency: ω = √(k/m) = √(200 / 0.50) = √(400) = 20.0 rad/s.
  3. Part (a): The maximum velocity occurs at equilibrium and is given by v_max = A ω.
  4. Calculate: v_max = 0.10 m · 20.0 rad/s = 2.00 m/s.
  5. Part (b): Use the velocity-displacement formula: v = ±ω√(A² - x²).
  6. Substitute x = 6.0 cm = 0.06 m: v = ±20.0 √(0.10² - 0.06²).
  7. Simplify terms: v = ±20.0 √(0.0100 - 0.0036) = ±20.0 √(0.0064).
  8. Calculate root: √(0.0064) = 0.08 m.
  9. Multiply: v = ±20.0 · 0.08 = ±1.60 m/s.
  10. The maximum velocity is 2.00 m/s, and the velocity at 6.0 cm is ±1.60 m/s.

Answer: v_max = 2.00 m/s, v(0.06) = ±1.60 m/s

An object in SHM has a maximum speed of 3.0 m/s and a maximum acceleration of 18.0 m/s². Find the amplitude and the angular frequency of the oscillation.
  1. Recall the formulas for maximum speed and acceleration: v_max = Aω and a_max = Aω².
  2. We are given v_max = 3.0 m/s and a_max = 18.0 m/s².
  3. Set up a ratio to find angular frequency ω: a_max / v_max = (Aω²) / (Aω) = ω.
  4. Calculate: ω = 18.0 / 3.0 = 6.0 rad/s.
  5. Now use the ω value to solve for Amplitude A: v_max = Aω ⇒ A = v_max / ω.
  6. Calculate: A = 3.0 / 6.0 = 0.50 meters.
  7. The angular frequency is 6.0 rad/s and the amplitude is 0.50 m.

Answer: ω = 6.0 rad/s, A = 0.50 m

Common Mistakes

  • Assuming velocity is max at max displacement: Confusing maximum distance with maximum speed. Velocity is actually zero at the extreme edges.
  • Sign errors in velocity: Forgetting the negative sign in v(t) = -Aω sin(ωt + φ), which is essential for preserving the phase relations.
  • Confusing angular frequency (ω) and frequency (f): Forgetting that ω = 2πf, leading to a 2π error in maximum speed calculations.
  • Ignoring double signs: Forgetting that v = ±ω√(A² - x²) has two signs, representing motion in both directions at any position x.

Practice Questions

1. Derive the velocity-displacement equation v(x) = ±ω√(A² - x²) from the principle of conservation of mechanical energy.

The total mechanical energy in SHM is constant: E = KE + PE = 1/2 m v² + 1/2 k x². At maximum displacement (x = A), the velocity is zero, so the total energy is purely potential: E = 1/2 k A². Equating these gives: 1/2 m v² + 1/2 k x² = 1/2 k A². Cancel the 1/2 terms: m v² + k x² = k A². Subtract k x² from both sides: m v² = k(A² - x²). Divide by m: v² = (k/m)(A² - x²). Since ω² = k/m, we get v² = ω²(A² - x²). Taking the square root gives v = ±ω√(A² - x²).

2. Explain the phase relationships between displacement, velocity, and acceleration vectors in Simple Harmonic Motion.

In SHM, velocity leads displacement by a phase of π/2 radians (90°), and acceleration leads velocity by another π/2 radians (making acceleration π radians or 180° out of phase with displacement). This means: (1) When displacement is zero (at equilibrium), velocity is at its maximum, and acceleration is zero. (2) When displacement is at positive amplitude (+A), velocity is zero, and acceleration is at its maximum negative value (-Aω²). (3) When displacement is at negative amplitude (-A), velocity is zero, and acceleration is at its maximum positive value (+Aω²).

3. Describe the significance of the phase space orbit (velocity vs. displacement) of a simple harmonic oscillator. What does its shape represent?

A phase space plot maps velocity (y-axis) against displacement (x-axis) as the system evolves. For SHM, the equation is x²/A² + v²/v_max² = 1. This is the equation of an ellipse. A closed elliptical path represents a stable, periodic, non-dissipative oscillation where total energy is conserved. The area of the ellipse is proportional to the total energy. If there were damping, the trajectory would spiral inward toward the origin, showing energy loss.

4. How does changing the mass (m) of a spring-mass system affect its maximum velocity if the initial amplitude is kept constant?

The maximum velocity is given by v_max = Aω, where ω = √(k/m). If amplitude A and spring constant k are kept constant, then v_max = A √(k/m) ∝ 1/√m. Therefore, increasing the mass decreases the angular frequency, which in turn reduces the maximum velocity of the system. Specifically, quadrupling the mass cuts the maximum velocity in half.

Frequently Asked Questions

What is velocity in Simple Harmonic Motion?

Velocity is the rate of change of position of the oscillating object with respect to time. It describes how fast and in what direction the object is moving at any given moment.

What is the standard formula for velocity in SHM?

As a function of time, v(t) = -Aω sin(ωt + φ). As a function of position, v(x) = ±ω√(A² - x²), where A is amplitude, ω is angular frequency, and φ is the phase constant.

Where is velocity maximum in SHM?

Velocity is maximum when the object passes through the equilibrium position (x = 0). At this point, the displacement is zero, and all potential energy has converted into kinetic energy.

Where is velocity zero in SHM?

Velocity is zero at the extreme positions or turning points (x = +A and x = -A), where the oscillator temporarily stops to reverse direction.

What is the phase difference between displacement and velocity in SHM?

Velocity leads displacement by π/2 radians (or 90°). This represents a quarter-cycle phase shift, meaning velocity is zero when displacement is maximum, and vice versa.

How is velocity related to energy in SHM?

The velocity of the mass directly determines its kinetic energy (K = 1/2 m v²). As the object oscillates, energy is continuously exchanged between kinetic energy and elastic potential energy (U = 1/2 k x²).

Can velocity be negative in SHM?

Yes. Velocity can be positive or negative depending on the direction of motion. The sign tells which way the oscillator is moving relative to the chosen positive direction.

How does amplitude affect maximum velocity in SHM?

Maximum velocity is vmax = Aω. If angular frequency stays the same, increasing amplitude increases the maximum velocity in direct proportion.

Why is velocity zero at the turning points?

At each turning point the object briefly stops before reversing direction. At that instant, all mechanical energy is potential energy, so velocity is zero.