Question

A 1.00-kg glider attached to a spring with a force constant of 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -3.00 cm

(a) Find the period of the glider's motion.


(b) Find the maximum values of its speed and acceleration.


(c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s.

Answer 1

(a) 0.126 s

(b) Maximum speed = 0.15 m/s

Maximum acceleration = 0.75 m/s2

(c) Position:
`x(t) = 0.03\cos(5t)`

Velocity:
`v(t) = -0.15\sin(5t)`

Acceleration:
`a(t) = -0.75\cos(5t)`

Step-by-step explanation:

(a) The motion is a simple harmonic motion. For a loaded spring, it's period, T, is given by

`T = 2\pi\sqrt{(m)/(k)}`

m and k are, respectively, the mass of the load and the force constant of the spring.

`T = 2\pi\sqrt{(1.00)/(25.0) = 0.4\pi = 0.126 \text{ s}}`

(b) The maximum speed is given by

`v_m=\omega A` and maximum acceleration by
`a_m=\omega^2 A`

where
`\omega` is the angular velocity and A is the amplitude or maximum displacement from the equilibrium or central point.

`\omega = \sqrt{(k)/(m)} = \sqrt{(25.0)/(1.00)}= 5 \text{ rad/s}`

A = 3 cm = 0.03 m

Hence,

`v_m=5* 0.03 = 0.15 \text{m/s}`

`a_m=5^2* 0.03 = 0.75 \text{m/s}^2`

(c) The general equation of SHM is

`x(t) = A\sin(\omega t \pm \phi)`

x(t) is the position at time, t

`\omega` is the angular velocity

T is the time

`\phi` is the phase difference which determines at what point the oscillation began. If it begins at the middle (which is almost practically impossible), it's value is 0. If it begins at any endpoint (which is most practical), its value is
`\pm\pi/2`, the sign depending on which side of the equilibrium point it begins from.

By trigonometry,
`\sin(\theta + \pi/2) = \cos(\theta)`.

Putting all values in x(t),

`x(t) = 0.03\cos(5t)`

Velocity is the time-derivative of position. Hence,

`v(t) = -0.15\sin(5t)`

Acceleration is the time-derivative of velocity. Hence,

`x(t) = -0.75\cos(5t)`