Question

A 0.500-kg mass on a spring has velocity as a function of timegiven by vx(t) = (3.60cm/s) sin[(4.71s-1)t -π/2].What is a) period? b) the amplitude? c) the maximumaccleration of the mass?

Answer 1

The period of the spring's oscillation is 0.212 seconds, the amplitude of the velocity function is 3.60 cm/s, and the maximum acceleration is 16.956 cm/s², which may need to be converted to m/s² for standard SI units.

Step-by-step explanation:

The question involves a mass-spring system undergoing simple harmonic motion (SHM), which can be described by sine or cosine functions. For this case, the given velocity function is vx(t) = (3.60 cm/s) sin[(4.71 s⁻¹)t - π/2].

Period of Oscillation

The period is the time it takes to complete one full cycle of motion and is given by the reciprocal of the coefficient of time in the argument of the sine function, which is 4.71 s⁻¹. Period T = 1 / 4.71 s⁻¹ = 0.212 s.

Amplitude of the Velocity

Amplitude is the maximum value of the velocity, which is given directly as 3.60 cm/s since that is the coefficient in front of the sine function.

Maximum Acceleration of the Mass

To find the maximum acceleration, we take the derivative of the velocity function to find acceleration a(t). The maximum acceleration occurs at the peak of the sine function, which is amax = 4.71 s⁻¹ × 3.60 cm/s = 16.956 cm/s². Remember to convert cm/s² to m/s² for SI units if needed.

Answer 2

The student's question regarding the simple harmonic motion of a spring-mass system involves finding the period, the amplitude, and the maximum acceleration from the provided velocity function. By analyzing the function, we can deduce that the period is related to the angular frequency, the amplitude is the coefficient in front of the sine function, and the maximum acceleration is the square of the angular frequency multiplied by the amplitude.

Step-by-step explanation:

The student asks about the characteristics of a mass-spring system undergoing simple harmonic motion (SHM), specifically regarding the period, amplitude, and maximum acceleration of the mass. Given the velocity function vx(t) = (3.60 cm/s) sin[(4.71 s-1)t - π/2], we can identify the properties of SHM.

The period (T) of SHM can be determined from the angular frequency ω (4.71 s-1), which is related to the period by the equation ω = 2π/T. Solving for T gives T = 2π/ω = 2π/4.71 s-1.

The amplitude (A) of the velocity function is 3.60 cm/s, as it is the coefficient of the sine function.

The maximum acceleration (amax) can be found using the relationship a = -ω2 x, where x is the amplitude of the displacement. If the velocity is at its maximum (amplitude), then the displacement x is zero, and hence acceleration is zero at that instant. However, the maximum acceleration occurs when the displacement is at its amplitude, and since we don't have the displacement function directly, we use the relation between maximum velocity and maximum acceleration in SHM: amax = ω2A. Substituting the given values, we have amax = (4.71 s-1)2 × 3.60 cm/s.