A 0.500kg⁡0.500\operatorname{kg} mass on a spring has velocity as a function of time given by vx(t)=−(3.60cm⁡/s)sin⁡[(4.7 rad/s)t−(π/2)]v_{x}(t)=-(3.60\operatorname{cm}/s)\sin[(4.7\text{ }rad/s)t-(\pi/2)]. What are the amplitude and the maximum acceleration of the mass?

Ch 14: Periodic Motion

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 14: Periodic Motion

Problema 20bc

Capitolo 14, Problema 20bc

A $0.500 kg$ mass on a spring has velocity as a function of time given by $v_{x} (t) = - (3.60 cm / s) \sin [(4.7 r a d / s) t - (π / 2)]$ . What are the amplitude and the maximum acceleration of the mass?

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To find the period (T) of the motion, use the angular frequency (ω) from the velocity function vx(t) = -(3.60 cm/s) sin[(4.71 rad/s)t - (π/2)]. The period is given by T = 2π/ω. Substitute ω = 4.71 rad/s into the formula to find T.

The amplitude (A) of the motion can be determined from the maximum velocity given in the function vx(t). The maximum velocity occurs when the sine function equals ±1, so the amplitude of the velocity is 3.60 cm/s. Use the relationship between maximum velocity and amplitude: vmax = ωA, where ω = 4.71 rad/s. Solve for A.

To find the maximum acceleration (amax), use the relationship amax = ω²A. You have already found ω and A in previous steps. Substitute these values into the formula to find amax.

The force constant (k) of the spring can be found using Hooke's Law and the relationship between angular frequency and mass: ω = √(k/m). Rearrange this formula to solve for k: k = mω². Use m = 0.500 kg and ω = 4.71 rad/s to find k.

Review the units and ensure consistency across all calculations. Convert units where necessary, such as converting cm/s to m/s for velocity, to maintain SI units throughout the problem.

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Simple Harmonic Motion

Simple Harmonic Motion (SHM) describes the oscillatory motion of systems like masses on springs, where the restoring force is proportional to displacement. The velocity function given indicates SHM, characterized by sinusoidal functions for displacement, velocity, and acceleration. Understanding SHM is crucial for determining the period, amplitude, and other properties of the system.

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Period of Oscillation

The period of oscillation is the time taken for one complete cycle of motion in SHM. It is inversely related to the angular frequency, ω, given in the velocity function. The period, T, can be calculated using T = 2π/ω, which helps in understanding the timing of the oscillatory motion and is essential for solving part (a) of the question.

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Hooke's Law states that the force exerted by a spring is proportional to its displacement, F = -kx, where k is the spring constant. The spring constant is crucial for determining the force characteristics of the spring system. It can be derived from the mass and angular frequency using k = mω², which is necessary for solving part (d) of the question.

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A 0.500kg0.500\(\operatorname{kg}\) mass on a spring has velocity as a function of time given by vx(t)=−(3.60cm/s)sin[(4.7 rad/s)t−(π/2)]v_{x}(t)=-(3.60\(\operatorname{cm}\)/s)\(\sin\)[(4.7\(\text{ }\)rad/s)t-(\(\pi\)/2)]. What are the amplitude and the maximum acceleration of the mass?

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Simple Harmonic Motion

Period of Oscillation

Hooke's Law and Spring Constant

A $0.500 kg$ mass on a spring has velocity as a function of time given by $v_{x} (t) = - (3.60 cm / s) \sin [(4.7 r a d / s) t - (π / 2)]$ . What are the amplitude and the maximum acceleration of the mass?