Examples

Example1:

A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find(a) the displacement of the particle,(b) its velocity, and(c) its acceleration.(d) Find the period and amplitude of the motion.

 

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Solution: a) The displacement as a function of time is x(t) = A cos(ωt + φ). Here ω = 2/s, φ = π/6, and A = 5 cm. The displacement at t = 0 is x(0) = (5 cm)cos(π/6) = 4.33 cm.

(b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.

(c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.(d) The period of the motion is T = π sec, and the amplitude is 5 cm.

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Example 2:

A mass-spring system oscillates with an amplitude of 3.5

cm. If the force constant of the spring of 250 N/m and the

mass is 0.5 kg, determine

(a) the mechanical energy of the system,

(b) the maximum speed of the mass, and

(c) the maximum acceleration .

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Solution:

(a) We have m = 0.5 kg, A = 0.035 m, k = 250 N/m, ω2 = k/m =

500/s2, ω = 22.36/s.

The mechanical energy of the system is E = (1/2)kA2 = 0.153 J.

(b) The maximum speed of the mass is vmax = ωA = 0.78 m/s.

(c) The maximum acceleration is amax = ω2A = 17.5 m/s2.

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Example 3:

A 20 g particle moves in simple harmonic motion with a frequency of 3 oscillations per second and an amplitude of 5cm.

(a)Through what total distance does the particle move during one cycle of its motion?

(b)(b) What is its maximum speed? Where does that occur?

(c)(c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur?

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Solution:

(a)The total distance d the particle moves during one cycle is from x = -A to x = +A and back to x = -A, so d = 4A = 20 cm.

(b) The maximum speed of the particle is vmax = ωA = 2πfA = 2π 15 cm/s = 0.94 m/s. The particle has maximum speed when it passes through the equilibrium position.

© The maximum acceleration of the particle is amax = ω2A = (2πf)2A = 17.8 m/s2. The particle has maximum acceleration at the turning points, where it has maximum displacement.

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EX 4: A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless air track.

(A) Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3.00 cm.

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1- An oscillator consists of a block of mass 0.50 kg connected to a spring. When set into oscillation with amplitude 35 cm, it is observed to repeat its motion every 0.50 s. The maximum speed is : (a) 4.4 m/s ,(b) 44.0 m/s ,( c) 44.0 m/s

2- A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. The max speed of the motion equal:(a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s

3- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant of the spring is, (a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m

4- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the block is, (a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg

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