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A body of mass \(m = 2\text{ kg} \) is oscillating on a spring. The position in meters of the body as a function of time \(t\) can be expressed as

\[ x(t) = A\cos{(2\pi ft)} \text{ m} .\]

If the phase of \( x(t)\) is shifted by \(\delta = \frac{8 \pi}{5} \text{ rad},\) the time delay of its cycle is \(KT \text{ s},\) where \(T\) is the period of the oscillation. What is the constant \(K?\)

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As illustrated in the above diagram, a body of mass \(m = 4 \text{ kg} \) attached to a spring with spring constant \( k = 16 \text{ N/m}\) is oscillating on a frictionless floor. Initially at \(t=0,\) the position of the body was \( x(t=0) = 4 \text{ m } \) and the velocity was \( v(t=0) = 0 \text{ m/s }.\) What is the amplitude of the oscillation?

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A body of mass \(m = 2 \text{ kg} \) is oscillating on a spring. The position of the body is given as \( x(t) = 4 \cos{(30 t)} \text{ m}, \) where \(x \) denotes the difference in the length of the spring from its original length. What is the period of the body?

Assume that \(\pi = 3 \).

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