Question

A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8.0 ✕ 103 m/s2. The phase constant is -π/2 rad. (a) Write an equation for the force on the particle as a function of time. (Use the following as necessary: t.) F = N (b) What is the period of the motion? s (c) What is the maximum speed of the particle? m/s (d) What is the total mechanical energy of this simple harmonic oscillator? J

Answer 1

Answer:

Explanation:

Given:

Mass, m = 30 g

= 0.03 kg

Amplitude, A = 2.0 ✕ 10^-3 m

Maximum acceleration, am = 8.0 ✕ 10^3 m/s2

Phase constant, phil = -π/2 rad

Displacement, x = A cos(wt + phil)

dx/dt = v = -Aw sin(wt + phil)

dv/dt = a = -Aw^2 cos(wt + phil)

Force, f = mass × acceleration

F = -0.03 × Aw^2 cos(wt + phil)

B.

At am,

a = Aw^2

8.0 ✕ 10^3 = 2.0 ✕ 10^-3 × w^2

w = sqrt(4 × 10^6)

= 2 × 10^3 rad/s

But w = 2pi/T

Where T = period

T = 2pi/2000

= 3.142 × 10^-3 s

= 0.00314 s

C.

Velocity = Aw

= 2.0 ✕ 10^-3 × 2000

= 4 m/s

D.

Total mechanical energy = kinetic energy + potential energy

= 1/2mv^2 + mgx

= 0.03 × [1/2 × 4^2 + (9.8 × (2 × 10^-3 × cos(2π - π/2)))]

= 0.03 × (8 + 0)

= 0.24 J

Answer 2

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

$F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)$

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

$\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s$

(a).

$F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})$

(b). $T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s$

(c). $v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}$

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

$E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J$