Question

The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers to three significant figures. Determine:
the amplitude
2cm
the period
.628s
the spring constant
5N/m
the phase constant
-.785rad
the initial coordinate of the mass
1.41cm
the initial velocity
14.1 cm/s
maximum speed 20 cm/s
what's the total energy?

Answer 1

The mass of the system is $m=50 g=0.05 kg$, and its position at time t is given by
$x=(2.0 cm)cos(10t-\frac{\pi}{4})$
which corresponds to the equation of the harmonic motion:
$x(t)=A cos (\omega t-\phi) where A is the amplitude, [tex]\omega$ is the angular frequency and $\phi$ is the phase constant.

1) Amplitude of the motion: this is given by A in the equation above, so in this case the amplitude is A=2.0 cm.

2) Period: the angular frequency of this system is the factor in front of t in the cosine, so $\omega=10 rad/s$. The period is related to the angular frequency by
$T=\frac{2 \pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

3) Spring constant: the angular frequency is related to the spring constant by
$\omega=\sqrt{\frac{k}{m}}$
where m is the mass of the system. By rearranging the equation, we find
$k=\omega^2 m=(10 rad/s)^2(0.05 kg)=5 N/m$

4) The phase constant is the second factor in the argument of the cosine, so in this case:
$\phi=-\frac{\pi}{4}=-0.785 rad$

5) Initial coordinate of the mass: this can be found by using t=0 inside the equation. We get:
$x(0)=(2.0 cm)cos(10*0-0.785)=(2.0 cm)cos(-0.785 rad)=1.41 cm$

6) The velocity at time t is equal to the derivative of the position:
$v(t)=x'(t)=-\omega A sin(\omega t+\phi)$
In our case,
$v(t)=-(10 rad/s)(2.0 cm)(sin (10t-0.785 rad)$
and by substituting t=0, we find the initial velocity:
$v(0)=-(10 rad/s)(2.0 cm) sin(-0.785 rad)=14.1 cm/s$

7) Maximum speed: the maximum speed is the value of the velocity when the sine in the expression of v(t) is 1, so
$v_{max}=\omega A=(10 rad/s)(2.0cm)=20 cm/s$

8) The total energy of the system is equal to the maximum kinetic energy of the system (in fact, when the kinetic energy is maximum, the mass is passing through the equilibrium position, so the elastic potential energy is zero and all the energy is kinetic energy), and it is given by:
$E=\frac{1}{2}mv_{max}^2=\frac{1}{2}(0.05 kg)(0.20m/s)^2=0.001 J$