Related questions with answers

A mass is placed on a frictionless, horizontal table. A spring $(k=100 \mathrm{~N} / \mathrm{m})$ , which can be stretched or compressed, is placed on the table. A $5.00-\mathrm{kg}$ mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zéro. A student moves the mass out to $x=4.0 \mathrm{~cm}$ and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time $t=3.00 \mathrm{~s}$ .

A block of mass 3.00 kg is placed against a horizontal spring of constant k=875 N/m and pushed so the spring compresses by 0.0700 m. If the block is now released and the surface is frictionless, calculate the block's speed after leaving the spring.

A wave is modeled at time t = 0.00 s with a wave function that depends on position. The equation is

y(x) = (0.30 m)sin(6.28 m^{-1}x).

The wave travels a distance of 4.00 meters in 0.50 s in the positive x-direction. Write an equation for the wave as a function of position and time.

Question

A mass is placed on a frictionless, horizontal table. A spring (k = 100 N/m), which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x = 4.0 cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t = 3.00 s.

Solution

Verified

Step 1

1 of 3

First, since there are no non-conservative forces acting on the system, the amplitude $A$ of the movement will be equal to the initial displacement $\left(4.0 \text{ cm}\right)$ . We can also calculate the angular frequency $\omega$ with the given force constant $k$ and the mass $m$ . Once we have calculated $\omega$ , we continue to write the equation for the general position $x\left(t\right)$ of the object in simple harmonic motion. Then, we differentiate $x\left(t\right)$ in respect to time to obtain $v\left(t\right)$ . Finally, we differentiate $v\left(t\right)$ in respect to time to obtain $a\left(t\right)$ . Note that $\phi=0$ because the position is $4.0 \text{ cm}$ when $t=0$ .