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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

User SelimOber
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2 Answers

5 votes

Answer:

Step-by-step 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

User Mirel Vlad
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3.5k points
0 votes

Answer:

Step-by-step 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{(a_(max))/(A)}=\sqrt{(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(rad)/(s))^(2)cos(2000(rad)/(s) t - (\pi)/(2))=240N*cos(2000(rad)/(s) t - (\pi)/(2))

(b).
T=(2\pi)/(\omega)=(2\pi)/(2000)=3.14*10^(-13) s

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

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


E_(m)=E_(k)(v_(max))=(mv_(max)^(2))/(2)=(30*10^(-3)kg(4(m)/(s))^(2))/(2)=0.024J

User Cherusker
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