Textbook Question
The position of a 50 g oscillating mass is given by ๐(t) = (2.0 cm) cos (10 t - ฯ/4), where t is in s. Determine: The initial conditions.
Ch 15: Oscillations
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 15, Problem 17h
The position of a 50 g oscillating mass is given by ๐(t) = (2.0 cm) cos (10 t - ฯ/4), where t is in s. Determine: The velocity at t = 0.40 s.
Verified step by step guidance1
Step 1: Recall the relationship between position and velocity in simple harmonic motion. Velocity is the time derivative of the position function. Start by differentiating the given position function ๐(t) = (2.0 cm) cos(10 t โ ฯ/4) with respect to time t.
Step 2: Apply the derivative rule for cosine: d/dt[cos(ฮธ)] = -sin(ฮธ) * dฮธ/dt. Here, ฮธ = (10 t โ ฯ/4), so dฮธ/dt = 10. The derivative of ๐(t) becomes v(t) = - (2.0 cm) * sin(10 t โ ฯ/4) * 10.
Step 3: Simplify the expression for velocity: v(t) = -20 cm/s * sin(10 t โ ฯ/4). This is the general formula for the velocity of the oscillating mass.
Step 4: Substitute t = 0.40 s into the velocity equation. Calculate the argument of the sine function: ฮธ = 10 * 0.40 โ ฯ/4.
Step 5: Evaluate sin(ฮธ) using the calculated ฮธ value, and multiply by -20 cm/s to find the velocity at t = 0.40 s. Ensure the units are consistent throughout the calculation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Simple Harmonic Motion (SHM)
Simple Harmonic Motion is a type of periodic motion where an object oscillates around an equilibrium position. The motion can be described by sinusoidal functions, such as sine or cosine, which represent the position of the object as a function of time. In this case, the mass oscillates with a specific amplitude and angular frequency, which are key to determining its velocity and other properties.
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Simple Harmonic Motion of Pendulums
Velocity in SHM
In Simple Harmonic Motion, the velocity of an oscillating object can be derived from its position function. The velocity is the time derivative of the position function, indicating how fast the object is moving and in which direction. For the given position function, differentiating it with respect to time will yield the velocity expression, which can then be evaluated at any specific time.
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Angular Frequency
Angular frequency, denoted by ฯ, is a measure of how quickly an object oscillates in radians per second. It is related to the frequency of oscillation and is crucial for determining the motion characteristics of oscillating systems. In the provided equation, the angular frequency is represented by the coefficient of t in the cosine function, which influences both the speed and the period of the oscillation.
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Related Practice
Textbook Question
A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vx = -30 cm/s. Determine: The position at t = 0.40 s.
Textbook Question
A 1.0 kg block is attached to a spring with spring constant 16 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40 cm/s. What are The block's speed at the point where ๐ = (ยฝ)A?
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Textbook Question
A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vโ = -30 cm/s. Determine: The total energy.
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Textbook Question
A 500 g air-track glider moving at 0.50 m/s collides with a horizontal spring whose opposite end is anchored to the end of the track. Measurements show that the glider is in contact with the spring for 1.5 s before it rebounds. What is the maximum compression of the spring?
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Textbook Question
A 500 g air-track glider moving at 0.50 m/s collides with a horizontal spring whose opposite end is anchored to the end of the track. Measurements show that the glider is in contact with the spring for 1.5 s before it rebounds. What is the value of the spring constant?