A spring with an 8 kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 5 m/s. Find the position (in m) of the mass at any time t.x(t) = _______ m x(t) = 1.5811sin(√10t) m --- this was marked wrong. Please help **Problem: Spring-Mass System**A spring with an 8 kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 5 m/s. Find the position (in meters) of the mass at any time $ t $.**Solution:**\[ x(t) = 1.5811 \sin(\sqrt{10} \, t) \, \text{m} \]**Explanation:**This formula represents the position $ x(t) $ of the mass at any given time $ t $. The position is modeled as a sinusoidal function due to the harmonic motion of the spring-mass system. The coefficient (1.5811) indicates the amplitude, which is the maximum displacement from the equilibrium position, while $ \sqrt{10} $ represents the angular frequency, determining how quickly the oscillations occur.

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A spring with an 8 kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 5 m/s. Find the position (in m) of the mass at any time t. x(t) = 1.5811 sin(√10 t) X m

A spring with an 8 kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 5 m/s. Find the position (in m) of the mass at any time t. x(t) = 1.5811 sin(√10 t) X m

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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x(t) = _______ m

x(t) = 1.5811sin(√10t) m --- this was marked wrong. Please help

**Problem: Spring-Mass System**

A spring with an 8 kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 5 m/s. Find the position (in meters) of the mass at any time $ t $.

**Solution:**

\[
x(t) = 1.5811 \sin(\sqrt{10} \, t) \, \text{m}
\]

**Explanation:**

This formula represents the position $ x(t) $ of the mass at any given time $ t $. The position is modeled as a sinusoidal function due to the harmonic motion of the spring-mass system. The coefficient (1.5811) indicates the amplitude, which is the maximum displacement from the equilibrium position, while $ \sqrt{10} $ represents the angular frequency, determining how quickly the oscillations occur.

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