174k views
4 votes
a particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the ori- gin, at t 5 0 and moves to the right. the amplitude of its motion is 2.00 cm, and the frequency is 1.50 hz. (a) find an expression for the position of the particle as a function of time. determine (b) the maximum speed of the particle and (c) the earliest time (t . 0) at which the particle has this speed. find (d) the maxi- mum positive acceleration of the particle and (e) the earliest time (t . 0) at which the particle has this accel- eration. (f) find the total distance traveled by the par- ticle between t 5 0 and t 5 1.00 s.

User HamidTB
by
8.2k points

1 Answer

4 votes

(a) The position of the particle as a function of time is given by:

x(t) = A cos(2πft)

where A is the amplitude (2.00 cm), f is the frequency (1.50 Hz), and cos is the cosine function.

Substituting the given values, we get:

x(t) = 2.00 cos(3πt)

(b) The maximum speed of the particle occurs at the equilibrium position, where the displacement is zero. At this point, the velocity is maximum and is given by:

vmax = Aω

where ω is the angular frequency and is equal to 2πf. Substituting the given values, we get:

vmax = 2.00 × 2π × 1.50 = 18.85 cm/s

(c) The earliest time at which the particle has this speed is when it passes through the equilibrium position. This happens at t = 0, so the earliest time is t = 0.

(d) The maximum positive acceleration of the particle occurs at the ends of its motion, where the displacement is maximum. At these points, the acceleration is given by:

amax = Aω^2

Substituting the given values, we get:

amax = 2.00 × (2π × 1.50)^2 = 282.74 cm/s^2

(e) The earliest time at which the particle has this acceleration is when it reaches the maximum displacement. This happens at t = 1/4T, where T is the period of the motion. The period is given by:

T = 1/f = 2/3 s

So, t = 1/4T = 1/4 × 2/3 = 0.33 s

(f) The total distance traveled by the particle between t = 0 and t = 1.00 s is equal to one complete cycle of its motion. The distance traveled in one complete cycle is equal to four times the amplitude, or:

4A = 8.00 cm

Therefore, the total distance traveled is:

8.00

User PyroAVR
by
9.2k points