Photo AI

A particle is moving along the -x-axis in simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2015 - Paper 1

Question icon

Question 13

A-particle-is-moving-along-the--x-axis-in-simple-harmonic-motion-HSC-SSCE Mathematics Extension 1-Question 13-2015-Paper 1.png

A particle is moving along the -x-axis in simple harmonic motion. The displacement of the particle is x metres and its velocity is v m s⁻¹. The parabola below shows ... show full transcript

Worked Solution & Example Answer:A particle is moving along the -x-axis in simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2015 - Paper 1

Step 1

For what value(s) of x is the particle at rest?

96%

114 rated

Answer

The particle is at rest when its velocity v is zero. From the equation v² = n²(a² - (x - c)²), setting v² to zero gives:

0=n2(a2(xc)2)0 = n²(a² - (x - c)²)

Solving for x results in:

(xc)2=a2(x - c)² = a²

Taking the square root of both sides gives:

xc=±ax - c = ±a

Thus, the values of x where the particle is at rest are:

x=c+ax = c + a and x=cax = c - a.

Step 2

What is the maximum speed of the particle?

99%

104 rated

Answer

The maximum speed occurs when the displacement x is zero, located at the maximum value of v² on the graph. From the equation for v², substituting x = c results in:

v2=n2(a2(cc)2)=n2a2v² = n²(a² - (c - c)²) = n²a²

Thus, the maximum speed v is given by:

v=nav = n a.

Step 3

What are the values of a, c and n?

96%

101 rated

Answer

The equation provided indicates the form of simple harmonic motion, where:

  • a is the amplitude of the motion,

  • c is the equilibrium position (the x-coordinate of the vertex of the parabola),

  • n relates to the angular frequency of the motion. Thus:

  • a corresponds to half the width of the parabola at the maximum height,

  • c is the x-value at the vertex of the parabola,

  • n is derived from the coefficient that relates to the speed at maximum displacement.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;