A mass on a spring has an amplitude of 6.0 cm and a period of 2.0 s. If it is released from its maximum displacement at t=0, what are its position and velocity at t=5.2 s?

here's the solution for the spring problem:

Given:
Amplitude, $A = 6.0 \text{ cm} = 0.06 \text{ m}$
Period, $T = 2.0 \text{ s}$
Time, $t = 5.2 \text{ s}$

Step 1: Calculate the angular frequency ($\omega$).
The angular frequency is related to the period by the formula:

$$\omega = \frac{2\pi}{T}$$

Substitute the given period:

$$\omega = \frac{2\pi}{2.0 \text{ s}} = \pi \text{ rad/s}$$

Step 2: Compute the position after 5.2 s.
Since the mass is released from its maximum displacement ($x=A$) at $t=0$, the displacement equation is:

$$x(t) = A \cos(\omega t)$$

Substitute the values for $A$ and $\omega$:

$$x(t) = 0.06 \cos(\pi t)$$

Now, calculate the position at $t = 5.2 \text{ s}$:

$$x(5.2) = 0.06 \cos(\pi \times 5.2)$$ $$x(5.2) = 0.06 \cos(5.2\pi)$$ $$x(5.2) \approx 0.06 \times (-0.8090)$$ $$x(5.2) \approx -0.04854 \text{ m}$$

The position after 5.2 s is $\boxed{\text{-0.0485 m}}$.

Step 3: Compute the velocity after 5.2 s.
The velocity is the derivative of the displacement with respect to time:

$$v(t) = \frac{d}{dt} (A \cos(\omega t)) = -A\omega \sin(\omega t)$$

Substitute the values for $A$ and $\omega$:

$$v(t) = -0.06\pi \sin(\pi t)$$

Now, calculate the velocity at $t = 5.2 \text{ s}$:

$$v(5.2) = -0.06\pi \sin(\pi \times 5.2)$$ $$v(5.2) = -0.06\pi \sin(5.2\pi)$$ $$v(5.2) \approx -0.06\pi \times (-0.5878)$$ $$v(5.2) \approx 0.1107 \text{ m/s}$$

The velocity after 5.2 s is $\boxed{\text{0.111 m/s}}$.

Step 4: Calculate the maximum velocity.
The maximum velocity ($v_{max}$) for Simple Harmonic Motion is given by:

$$v_{max} = A\omega$$

Substitute the values for $A$ and $\omega$:

$$v_{max} = (0.06 \text{ m}) \times (\pi \text{ rad/s})$$ $$$$