12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres. The displacement is given by $x = 2 ext{sin } 3t$, where $t$ is time in seconds. The motion starts when $t = 0$. (a)(i) What is the total distance travelled by the particle when it first returns to the origin? (ii) What is the acceleration of the particle when it is first at rest? (b) The region bounded by $y = ext{cos } 4x$ and the $x$-axis, between $x = 0$ and $x = rac{ ext{π}}{2}$, is rotated about the $x$-axis to form a solid. Find the volume of the solid. (c) A particle moves along a straight line with displacement $x$ m and velocity $v$ m s$^{-1}$. The acceleration of the particle is given by $rac{d^2x}{dt^2} = 2 - rac{x}{2}$. Given that $v = 4$ when $x = 0$, express $v^2$ in terms of $x$.

Photo AI

12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1

Question 12

12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres. The displacement is given by $x = 2 ext{sin } 3t$, where $t$ is t... show full transcript

Worked Solution & Example Answer:12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1

Step 1

What is the total distance travelled by the particle when it first returns to the origin?

96%

114 rated

Answer

To find the total distance travelled by a particle in simple harmonic motion, we need to determine the period of the motion. The displacement function is given by

$x = 2 ext{sin } 3t$ .

To find the period $T$ , we use the coefficient of $t$ inside the sine function:

$T = rac{2 ext{π}}{ ext{frequency}} = rac{2 ext{π}}{3}$ .

Thus, the time at which the particle first returns to the origin (where $x = 0$ ) is at the end of the first half of the period, or $rac{T}{2} = rac{ ext{π}}{3}$ .

The total distance travelled to return to the origin is twice the amplitude of the motion, hence:

$$ ext{Distance} = 2 imes 2 = 4 ext{ metres}.$

Step 2

What is the acceleration of the particle when it is first at rest?

99%

104 rated

Answer

The particle is at rest when its velocity is zero. The velocity is derived from the displacement function:

$v = rac{dx}{dt} = 2 imes 3 ext{cos}(3t) = 6 ext{cos}(3t).$

Setting $v = 0$ gives:

$6 ext{cos}(3t) = 0 \Rightarrow ext{cos}(3t) = 0.$

This occurs at:

$3t = rac{ ext{π}}{2}, \Rightarrow t = rac{ ext{π}}{6}.$

To find the acceleration, we differentiate the velocity:

$a = rac{dv}{dt} = -6 imes 3 ext{sin}(3t) = -18 ext{sin}(3t).$

Substituting $t = rac{ ext{π}}{6}$ gives:

$$a = -18 ext{sin}igg(3 imes rac{ ext{π}}{6}igg) = -18 ext{sin}igg(rac{ ext{π}}{2}igg) = -18 imes 1 = -18 ext{ m/s}^2.$

Step 3

Find the volume of the solid.

96%

101 rated

Answer

To find the volume of the solid formed by rotating the region bounded by $y = ext{cos } 4x$ about the $x$ -axis, we use the formula for the volume of revolution:

$V = ext{π} \int_{a}^{b} [f(x)]^2 \, dx,$

where $f(x) = ext{cos } 4x$ , $a = 0$ , and $b = rac{ ext{π}}{2}$ . Thus,

$V = ext{π} \int_{0}^{ rac{ ext{π}}{2}} ( ext{cos } 4x)^2 \, dx.$

Using the double angle formula, we have:

$( ext{cos } 4x)^2 = rac{1 + ext{cos } 8x}{2}.$

Therefore,

$V = ext{π} \int_{0}^{ rac{ ext{π}}{2}} rac{1 + ext{cos } 8x}{2} \, dx = rac{ ext{π}}{2} \int_{0}^{ rac{ ext{π}}{2}} (1 + ext{cos } 8x) \, dx.$

Calculating the integral gives:

$$V = rac{ ext{π}}{2} \left[x + rac{1}{8} ext{sin } 8x \right]_{0}^{rac{ ext{π}}{2}} = rac{ ext{π}}{2} \left[rac{ ext{π}}{2} + 0 - (0 + 0)\right] = rac{ ext{π}^2}{4}.$

Step 4

Given that $v = 4$ when $x = 0$, express $v^2$ in terms of $x$.

98%

120 rated

Answer

To express $v^2$ in terms of $x$ , we start from the equation for acceleration:

$rac{d^2x}{dt^2} = 2 - rac{x}{2}.$

We have: $rac{dv}{dt} = rac{dx}{dt} \times rac{dv}{dx} \Rightarrow v \frac{dv}{dx} = 2 - rac{x}{2}.$

This gives us a first order separable equation:

$v \frac{dv}{dx} = 2 - rac{x}{2}.$

Integrating both sides:

$\int v \, dv = \int (2 - \frac{x}{2}) \, dx$

which results in

$\frac{1}{2}v^2 = 2x - \frac{1}{4}x^2 + C.$

Using the condition $v = 4$ when $x = 0$ :

$\frac{1}{2}(4)^2 = 0 + C \Rightarrow C = 8.$

Thus,

$\frac{1}{2}v^2 = 2x - \frac{1}{4}x^2 + 8.$

Finally, multiplying by 2 yields:

$v^2 = 4x - \frac{1}{2}x^2 + 16.$

12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1

Worked Solution & Example Answer:12. A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1

What is the total distance travelled by the particle when it first returns to the origin?

What is the acceleration of the particle when it is first at rest?

Find the volume of the solid.

Given that $v = 4$ when $x = 0$, express $v^2$ in terms of $x$.

Join the SSCE students using SimpleStudy...