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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
Step 1
Answer
To find the total distance traveled by the particle when it first returns to the origin, we need to determine the period of the motion. The displacement function is given by:
The particle returns to the origin when:
ightarrow 2 ext{ sin }(3t) = 0 ightarrow ext{ sin }(3t) = 0. $$ This occurs when: $$ 3t = n ext{π} ext{ for } n ext{ an integer.} $$ The first return happens at: $$ t = rac{ ext{π}}{3}. $$ The motion is periodic with a period of: $$ T = rac{2 ext{π}}{3}. $$ The distance traveled during one complete cycle is equal to twice the amplitude - when the particle moves from the maximum displacement to the minimum and back. Therefore, the total distance is: $$ ext{Total Distance} = 2 imes 2 = 4 ext{ metres}. $$Step 2
Answer
The particle is first at rest when its velocity is zero.
The velocity function can be derived from the displacement:
v(t) = rac{dx}{dt} = rac{d}{dt} ig( 2 ext{ sin }(3t) ig) = 6 ext{ cos }(3t).
To find when the particle is at rest:
ightarrow ext{ cos }(3t) = 0. $$ This occurs for: $$ 3t = rac{ ext{π}}{2} + n ext{π} ightarrow t = rac{ ext{π}}{6} + rac{n ext{π}}{3}. $$ At $t = rac{ ext{π}}{6}$, the acceleration can be found using: $$ a(t) = rac{d^2x}{dt^2} = rac{d}{dt}(6 ext{ cos }(3t)) = -18 ext{ sin }(3t). $$ Calculating the acceleration at $t = rac{ ext{π}}{6}$: $$ aigg(rac{ ext{π}}{6}igg) = -18 ext{ sin }igg(3 imes rac{ ext{π}}{6}igg) = -18 ext{ sin }(rac{ ext{π}}{2}) = -18 ext{ m/s}^2. $$Step 3
Answer
To find the volume of the solid formed by rotating the region bounded by and the -axis from to x = rac{ ext{π}}{2}, we use the disk method:
Thus:
= ext{π} imes igg( ext{integral}_{0}^{rac{ ext{π}}{2}} rac{1 + ext{cos}(8x)}{2} dx igg) \ = ext{π} imes igg( rac{1}{2}igg[xigg]_{0}^{rac{ ext{π}}{2}} + rac{1}{16}igg[ ext{sin}(8x)igg]_{0}^{rac{ ext{π}}{2}} igg) \ = ext{π} imes igg( rac{ ext{π}}{4} + 0 igg) \ = rac{ ext{π}^2}{4}. $$Step 4
Answer
Using the given relationship:
a = rac{dv}{dt} = rac{d^2x}{dt^2},
we can relate velocity and acceleration. Since:
v = rac{dx}{dt},
we rewrite it as:
rac{dv}{dx} = rac{d^2x}{dt^2} rac{1}{v}.
Replacing the acceleration:
v rac{dv}{dx} = 2 - rac{x}{2}.
To separate the variables, we integrate:
ext{integral}(v dv) = ext{integral}igg(2 - rac{x}{2}igg) dx.
This leads to:
rac{v^2}{2} = 2x - rac{x^2}{4} + C.
To find the constant , we use the condition that when :
ightarrow C = 8. $$ Thus: $$ rac{v^2}{2} = 2x - rac{x^2}{4} + 8 \ ightarrow v^2 = 4x - rac{x^2}{2} + 16. $$Report Improved Results
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