A 0.500kg⁡0.500\operatorname{kg}0.500kg 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)]vx(t)=−(3.60cm/s)sin[(4.7 rad/s)t−(π/2)]. What are the period and the force constant of the spring?

Ch 14: Periodic Motion

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

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Young & Freedman Calc 15th Edition

Ch 14: Periodic Motion

Problem 20ad

Chapter 14, Problem 20ad

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

Verified step by step guidance

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.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

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.

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.

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Spring Force (Hooke's Law)

Related Practice

Textbook Question

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

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Textbook Question

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

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Textbook Question

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?

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Textbook Question

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?

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Textbook Question

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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Textbook Question

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

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