Question

A 1.30-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 21.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.
(a) Find the force constant of the spring.
N/m

(b) Find the frequency of the oscillations.
Hz

(c) Find the maximum speed of the object.
m/s

(d) Where does this maximum speed occur?
x = ± m

(e) Find the maximum accel

Answer 1

Using Hooke's Law:
a)
$$
F=kX\\
k= \frac{1}{X}F\,
$$
Where, \(k\) is the force constant of the spring, \(F\) is the Force exerted up the spring and \(X\) is the displacement of the string with this Force:
$$
k= \frac{1}{0.002}21\,
$$
b)
$$
x=A\cos(\omega t-\phi)
$$
Where A is the maximum displacement, \(\omega\) is the angular frequency and \(\phi\) is the phase difference, take it as zero.
$$
\omega=\sqrt{\cfrac{k}{m}}
$$
Remember to use SI units for \(m\) (i.e. kg).

The frequency in hertz is
$$
\omega=2\pi f\\
f=\cfrac{\omega}{2\pi}
$$
c)
$$
v(t) = \frac{\mathrm{d} x}{\mathrm{d} t} = - A\omega \sin(\omega t-\varphi)
$$
The amplitude of this equation is the maximum speed.
d)
The maximum speed occurs with \(\sin(\omega t-\varphi)=1\). Which occurs, when \(\omega t -\varphi=\cfrac{(2n+1)\pi}{2}\) for \(n\in \mathbb{N}\). Take \(\varphi=0\) and solve for \(t\). What is the value of \(x\) at this \(t\)? This is where the maximum speed occurs.
e)
Acceleration is
$$
a(t) = \frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = - A \omega^2 \cos( \omega t-\varphi).
$$
The maximum acceleration is just the amplitude of this equation.