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 $x = rac{ ext{$ 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 $a = rac{d^{2}x}{dt^{2}}$. Given that $v = 4$ when $x = 0$, express $v^{2}$ in terms of $x$.

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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

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... show full transcript

Worked Solution & Example Answer: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?

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Answer

To determine the total distance travelled by the particle, we start with the displacement function given by

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

The particle returns to the origin when $x(t) = 0$ . The first instance occurs when:

$ext{sin}(3t) = 0$

This happens at the values of $t$ such that:

$3t = n ext{π}$

for $n = 0, 1, 2, ...$ . The smallest positive solution occurs at $n = 1$ :

$t = rac{ ext{π}}{3}.$

Next, we compute the distance travelled over one complete oscillation (from origin back to the next origin). The amplitude is $2$ , meaning the particle travels $2$ units to the maximum and $2$ units back to the origin, yielding a total distance of:

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

Step 2

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

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Answer

To find the acceleration, we first calculate the velocity by taking the derivative of the displacement function:

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

The particle is at rest when $v(t) = 0$ :

ightarrow ext{cos}(3t) = 0.$$ This gives us the smallest positive solution: $$3t = rac{ ext{π}}{2} ightarrow t = rac{ ext{π}}{6}.$$ To find the acceleration, we now calculate the second derivative of the displacement function: $$a(t) = rac{d^2x}{dt^2} = -6 imes 3 ext{ sin }(3t) = -18 ext{ sin }(3t).$$ Substituting $t = rac{ ext{π}}{6}$ into the acceleration function gives us: $$aigg( rac{ ext{π}}{6}igg) = -18 ext{ sin }igg(3 imes rac{ ext{π}}{6}igg) = -18 ext{ sin }igg( rac{ ext{π}}{2}igg) = -18.$$ Thus, the acceleration when the particle is first at rest is $-18 ext{ m/s}^{2}$.

Step 3

Find the volume of the solid.

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Answer

The volume $V$ of the solid formed by rotating the curve about the $x$ -axis is calculated using the formula:

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

where $f(x) = ext{cos} \, 4x$ , $a = 0$ , and $b = rac{ ext{π}}{2}$ . Therefore, we have:

$V = ext{π} imes ext{integral}_{0}^{ rac{ ext{π}}{2}} [ ext{cos}(4x)]^2 \, dx.$

Using the identity:

$ext{cos}^2 x = rac{1 + ext{cos}(2x)}{2},$

we can rewrite:

$V = ext{π} imes ext{integral}_{0}^{ rac{ ext{π}}{2}} rac{1 + ext{cos}(8x)}{2} \, dx = rac{ ext{π}}{2} igg[ x + rac{1}{8} ext{sin}(8x) igg]_{0}^{ rac{ ext{π}}{2}}.$

Substituting the limits yields:

$V = rac{ ext{π}}{2} igg[ rac{ ext{π}}{2} + 0 - (0 + 0)igg] = rac{ ext{π}^2}{4}.$

Thus, the volume of the solid is $rac{ ext{π}^2}{4}$ .

Step 4

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

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Answer

We start from the relation given for acceleration:

$rac{dv}{dt} = rac{d^{2}x}{dt^{2}} = rac{d}{dt}(v) = rac{dv}{dx} rac{dx}{dt} = v rac{dv}{dx}.$

Given that the acceleration is also expressed as:

$a = rac{dv}{dt} = - rac{x}{2},$

we equate the two expressions:

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

Separating variables gives us:

$2v \, dv = -x \, dx.$

Integrating both sides yields:

$v^{2} = - rac{x^{2}}{2} + C,$

where $C$ can be determined using the condition that $v = 4$ when $x = 0$ :

ightarrow C = 16.$$ Thus, we find: $$v^{2} = 16 - rac{x^{2}}{2}.$$

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: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 $x$.

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