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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.

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 \(g=10\text{ m/s}^2.\)

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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 \( t = 3 \text{ s } ? \)

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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 \(g=10\text{ m/s}^2.\)

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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 \( t = 5 \text{ s } ? \)

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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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