Answer to Question #194593 in Physics for Dave

Question

A 0.400 kg block is attached to a spring with spring constant 500.0 N/m and oscillates horizontally with an amplitude of 2.66 cm on a frictionless table. The x-axis is defined such that when the block is at x = 0, the spring is at its equilibrium length. At time t = 0.100 s, the block is at x = -1.49 cm and is moving in the positive x direction.

 

a) What is the period of the block's oscillation?

 

b) What is the block's speed at t = 0.100 s? (optional: enter this value, in m/s, in the blank below)

 

c) What is the block’s maximum speed?

 

d) What is the block's acceleration at t = 0.100 s?

Answer 1

a) The period of the block's oscillation:

$$T=2\pi\sqrt{m/k}=0.178\text{ s}.$$

b) The speed of the block at t=0.1 s (at x=-1.49 cm) can be found from conservation of energy. The potential energy is maximum at x=2.66 cm, then it gradually converts into kinetic energy, which is maximum at x=0:

$$E_{pm}=\frac12kA^2.$$

According to conservation of energy:

$$E_{pm}=E_p+E_k,\\\space\\ \frac12 kA^2=\frac12kx^2+\frac12mv^2,\\\space\\ v=\sqrt{\frac{k(A^2-x^2)}{m}}=0.779\text{ m/s}.$$

c) The maximum speed can be found from conservation of energy as well:

$$E_{km}=E_{pm},\\\space\\ \frac12mv^2_m=\frac12kA^2,\\\space\\ v_m=A\sqrt{k/m}=0.940\text{ m/s}.$$

d) Find equation for velocity:

$$a=\frac Fm=\frac{kx}{m}=18.625\text{ m/s}^2.$$