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Simple Harmonic Motion

A humbling fraction of physics boils down to direct application of simple harmonic motion, the description of oscillating objects. Learn the basis for springs, strings, and quantum fields.

Solutions to Simple Harmonic Motion

         

A \(40\text{ g}\) cube of edge length \(l=3\text{ cm}\) floats on water, oscillating up and down. Initially at \(t=0,\) the position of the body was \( x(t=0) = 5 \text{ cm } \) and the velocity was \( v(t=0) = 0 \text{ m/s },\) where \(x \) denotes the vertical difference in the position of the cube from the equilibrium point \(x_0.\) What is the velocity \(v(t)\) of the body at \( t = 2 \text{ s } ? \)

Assumptions and Details

  • Positive sign of \(x(t) \) means downward direction.
  • Ignore any friction.
  • The density of water is \( \rho = 1 \text{ g/cm}^3.\)
  • The gravitational acceleration is \(g=10\text{ m/s}^2.\)

As illustrated in the above diagram, a body of mass \(m = 1 \text{ kg} \) attached to a spring with spring constant \( k = 9 \text{ N/m}\) is oscillating on a frictionless floor. Initially at \(t=0,\) the position of the body was \( x(t=0) = 1 \text{ m } \) and the velocity was \( v(t=0) = 0 \text{ m/s },\) where \(x \) denotes the difference in the length of the spring from its original length.

What is the position of the body at \( t = 3 \text{ s } ? \)

A \(40\text{ g}\) cube of edge length \(l=5\text{ cm}\) floats on water, oscillating up and down. Initially at \(t=0,\) the position of the body was \( x(t=0) = 4 \text{ cm } \) and the velocity was \( v(t=0) = 0 \text{ m/s },\) where \(x \) denotes the vertical difference in the position of the cube from the equilibrium point \(x_0.\) What is \(x(t)\) at \( t = 2 \text{ s } ? \)

Positive sign of \(x(t) \) means downward direction.
Ignore any friction.
The density of water is \( \rho = 1 \text{ g/cm}^3.\)
The gravitational acceleration is \(g=10\text{ m/s}^2.\)

As illustrated in the above diagram, a body of mass \(m = 3 \text{ kg} \) attached to a spring with spring constant \( k = 75 \text{ N/m}\) is oscillating on a frictionless floor. Initially at \(t=0,\) the position of the body was \( x(t=0) = 1 \text{ m } \) and the velocity was \( v(t=0) = 0 \text{ m/s },\) where \(x \) denotes the difference in the length of the spring from its original length. What is the velocity of the body at \( t = 5 \text{ s } ? \)

A body of mass \(m = 6 \text{ kg} \) is oscillating on two identical springs with spring constant \( k = 48 \text{ N/m},\) as shown in the above diagram, where \(x \) denotes the downward difference from the equilibrium point \(x_0.\) Initially, the difference of the body was \( x(t=0) = 4 \text{ m } \) and the velocity was \( v(t=0) = 0 \text{ m/s }.\) What is the position of the body at \( t = 4 \text{ s } ? \)

Assumptions and Details

  • The gravitational acceleration is \( g = 10 \text{m/s}^2. \)
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