Skip to main content
Ch 14: Periodic Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 14: Periodic MotionProblema 20ad
Capitolo 14, Problema 20ad

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 period and the force constant of the spring?

Guida verificata passo dopo passo
1
To find the period (T) of the motion, use the angular frequency (ω) given in the velocity function. The formula for the period is T = 2π/ω. Here, ω = 4.71 rad/s, so substitute this value into the formula to find T.
The amplitude (A) of the motion can be determined from the velocity function. The maximum velocity occurs when the sine function equals ±1, which gives the amplitude of the velocity as 3.60 cm/s. Since velocity is the derivative of displacement, the amplitude of displacement can be found by considering the relationship between maximum velocity and amplitude: A = vmax/ω. Substitute vmax = 3.60 cm/s and ω = 4.71 rad/s to find A.
To find the maximum acceleration (amax), use the relationship between acceleration and displacement in simple harmonic motion: a(t) = -ω²x(t). The maximum acceleration occurs when the displacement is at its maximum amplitude. Therefore, amax = ω²A. Use the previously calculated amplitude and ω = 4.71 rad/s 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 = sqrt(k/m). Rearrange this formula to solve for k: k = mω². Substitute m = 0.500 kg and ω = 4.71 rad/s to find k.
Review the units and ensure all calculations are consistent. Convert units where necessary, such as converting cm to m for amplitude, to maintain SI units throughout the calculations.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
4m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) describes the oscillatory motion of systems like masses on springs, characterized by sinusoidal functions of time. The velocity function given, vx(t) = -(3.60 cm/s) sin[(4.71 rad/s)t - (pi/2)], indicates SHM, where the angular frequency, amplitude, and phase shift are key parameters. Understanding SHM is crucial for determining the period, amplitude, and other dynamic properties of the system.
Video consigliato:
Percorso guidato
07:52
Simple Harmonic Motion of Pendulums

Angular Frequency and Period

Angular frequency (ω) is a measure of how quickly an object oscillates in SHM, given in radians per second. It is related to the period (T), the time taken for one complete cycle, by the formula ω = 2π/T. In the given function, ω = 4.71 rad/s, allowing us to calculate the period using T = 2π/ω, which is essential for part (a) of the question.
Video consigliato:
Percorso guidato
05:08
Circumference, Period, and Frequency in UCM

Hooke's Law and Spring Constant

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 can be derived from the mass and angular frequency using the formula k = mω², where m is the mass of the object. This concept is vital for determining the force constant of the spring in part (d) of the question.
Video consigliato:
Percorso guidato
05:27
Spring Force (Hooke's Law)
Pratica correlata
Domanda del libro di testo

A 0.400-kg object undergoing SHM has ax = -1.80 m/s2 when x = 0.300 m. What is the time for one oscillation?

1781
views
1
rank
Domanda del libro di testo

For the oscillating object in Fig. E14.4, what is its maximum speed?

2163
views
Domanda del libro di testo

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from x = 0.090 m to x = -0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel from x = 0.090 m to x = -0.090 m?

1748
views
Domanda del libro di testo

Weighing Astronauts. This procedure has been used to 'weigh' astronauts in space: A 42.5-kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

1714
views
Domanda del libro di testo

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?

2484
views
Domanda del libro di testo

For the oscillating object in Fig. E14.4, what is its maximum acceleration?

2836
views