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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

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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 find the distance travelled by the particle when it first returns to the origin, we need to determine the period of the motion. The general formula for simple harmonic motion displacement is given by:

x(t)=Aextsin(heta)x(t) = A ext{ sin }( heta)

This implies that when the particle returns to the origin, the angle heta heta must equal 00 or extπ ext{π}, for the first complete return.

Given that x=2extsin(3t)x = 2 ext{ sin }(3t), we set:

2extsin(3t)=02 ext{ sin }(3t) = 0

The first return to the origin occurs at:

3t=0extorextπ3t = 0 ext{ or } ext{π}

Thus, the first return occurs at:

At this time, we calculate the total distance travelled. Within half a period:

  • From t=0t = 0 to t = rac{ ext{π}}{3}, the displacement goes from 00 to 22 m and back to 00 m.
  • Therefore, the distance is: 2+0=2+2=4extm|2| + |0| = 2 + 2 = 4 ext{ m}.

Step 2

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

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The particle is at rest when its velocity is zero. The velocity of the particle can be found by differentiating the displacement function:

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

Setting this equal to zero to find when the particle is at rest:

6 ext{ cos }(3t) = 0 \ \ ext{ implies } 3t = rac{ ext{π}}{2} \ \ ext{ or } 3t = rac{3 ext{π}}{2}

This gives:

t = rac{ ext{π}}{6} \ \ ext{ or } t = rac{ ext{π}}{2}

The first instance occurs at t = rac{ ext{π}}{6}. Now, substitute back to find the acceleration:

a(t) = rac{d^{2}x}{dt^{2}} = -18 ext{ sin }(3t).

Thus, when t = rac{ ext{π}}{6}:

$$aigg( 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 ext{ (1)} = -18 ext{ m/s}^{2}.$

Step 3

Find the volume of the solid.

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Answer

To find the volume of the solid formed by rotating the region bounded by y=extcos(4x)y = ext{cos} (4x), the xx-axis between x=0x = 0 and x = rac{ ext{π}}{2}, we use the disk method:

$$V = ext{π} ext{ } igg[ ext{ Area integral}igg] = ext{π} ext{ } igg( ext{Area under } y = ext{cos}(4x)igg) = ext{π} ext{ }igg( igg[ rac{1}{2} y^2 igg]_{0}^{ rac{ ext{π}}{2}} \ = ext{π} imes igg[ rac{1}{2}( ext{cos}(4 imes rac{ ext{π}}{2}))^2 - rac{1}{2}( ext{cos}(4 imes 0))^2\ = ext{π} imes igg[ rac{1}{2}(0^2) - rac{1}{2}(1^2) \ = - rac{ ext{π}}{2} \ \ V = ext{π} igg[ rac{1}{2}(1 - 0) = rac{ ext{π}}{2} ext{ } ( ext{Final positive volume})igg] = rac{ ext{π}}{2} ext{ m}^{3}.$

Step 4

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

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Answer

The acceleration is given by:

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

To find v2v^2 in terms of xx, we apply the chain rule:

rac{dv}{dt} = rac{dv}{dx} imes rac{dx}{dt}

This means:

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

Integrating:

rac{1}{2} v^2 = 2x - rac{x^2}{4} + C

Given that v=4v = 4 when x=0x = 0:

We can find CC. Therefore, C=8C = 8, giving:

v2=8+8=16v^2 = 8 + 8 = 16

Thus, the complete expression for v2v^2 is:

$$v^2 = 2x - rac{x^2}{4} + 16.$

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