Math Problem Statement

A weight is attached to a spring and reaches its equilibrium position (x = 0). It is then set in motion resulting in a displacement of x = 14 cos(t), where x is measured in centimeters and t is measured in seconds. What is the spring's velocity when t = 0 and when t = π/3?

Solution

The problem describes the motion of a weight attached to a spring, where the displacement is given by:

x(t)=14cos(t)x(t) = 14 \cos(t)

This is a simple harmonic motion, and the velocity function can be derived by differentiating the displacement x(t)x(t) with respect to time tt.

(a) Finding the spring's velocity when t=0t = 0 and t=π3t = \frac{\pi}{3}

1. Differentiate the displacement function:

The velocity v(t)v(t) is the derivative of x(t)x(t) with respect to tt: v(t)=dxdt=14sin(t)v(t) = \frac{dx}{dt} = -14 \sin(t)

2. (b) Velocity at t=0t = 0:

Substitute t=0t = 0 into the velocity function: v(0)=14sin(0)=0cm/secv(0) = -14 \sin(0) = 0 \, \text{cm/sec} This confirms the answer provided in the image, which is 0cm/sec0 \, \text{cm/sec}.

3. Velocity at t=π3t = \frac{\pi}{3}:

Substitute t=π3t = \frac{\pi}{3} into the velocity function: v(π3)=14sin(π3)v\left(\frac{\pi}{3}\right) = -14 \sin\left(\frac{\pi}{3}\right)

Since sin(π3)=32\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, we get: v(π3)=14×32=73cm/secv\left(\frac{\pi}{3}\right) = -14 \times \frac{\sqrt{3}}{2} = -7\sqrt{3} \, \text{cm/sec} Numerically, this is approximately: v(π3)7×1.732=12.1cm/secv\left(\frac{\pi}{3}\right) \approx -7 \times 1.732 = -12.1 \, \text{cm/sec}

Therefore, the velocity at t=π3t = \frac{\pi}{3} is approximately 12.1cm/sec-12.1 \, \text{cm/sec}.


Would you like more details on any specific part of the solution?

Here are 5 follow-up questions to consider:

  1. What is the maximum velocity of the spring's motion?
  2. How would the velocity change if the displacement function were x(t)=10cos(2t)x(t) = 10 \cos(2t)?
  3. What is the period of this spring's motion?
  4. How does changing the amplitude affect the velocity at different times?
  5. What is the spring's velocity at t=π/2t = \pi/2?

Tip: Remember, velocity is always the derivative of the displacement function when dealing with harmonic motion.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Harmonic Motion

Formulas

Displacement function x(t) = 14 cos(t)
Velocity function v(t) = dx/dt = -14 sin(t)

Theorems

Derivative of trigonometric functions
Simple Harmonic Motion

Suitable Grade Level

Grades 11-12