A particle moves in simple harmonic motion described by the equation
\( \ddot{x}=-9(x-4) . \)
Find the period and the central point of motion. (2 marks)
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A particle moves in simple harmonic motion described by the equation
\( \ddot{x}=-9(x-4) . \)
Find the period and the central point of motion. (2 marks)
\(\text{Period}\ = \dfrac{2 \pi}{3} \)
\(\text{Centre of Motion:}\ x=4 \)
\( \ddot{x}=-9(x-4) \)
\( \Rightarrow\ n=3,\ \ c=4 \)
\(\text{Period}\ = \dfrac{2 \pi}{3} \)
\(\text{Centre of Motion:}\ x=4 \)
The diagrams show two positions of a single piston in the cylinder chamber of a motorcycle. The piston moves vertically, in simple harmonic motion, between a maximum height of 0.17 metres and a minimum height of 0.05 metres.
The mass of the piston is 0.8 kg. The piston completes 40 cycles per second.
What is the resultant force on the piston, in newtons, that produces the maximum acceleration of the piston? Give your answer correct to the nearest newton. (3 marks)
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`text{3032 N}`
`text{Using}\ \ ddotx=-n^2(x-b)`
`text{Centre of motion}\ (b) = 0.11`
`text{Amplitude}\ (a) = (0.17-0.05)/2=0.06\ text{m}`
`text{Period}\ (T) = 1/40=0.025\ text{sec}`
`n=(2pi)/T=80pi`
`:.ddotx=-(80pi)^2(x-0.11)`
`ddotx_(max)\ text{occurs when}\ x=0.17\ text{or}\ 0.05`
`ddotx_(max)` | `=-(80pi)^2(0.17-0.11)` | |
`=384pi^2\ text{m s}^(-2)` |
`:.F_(max)` | `=mddotx` | |
`=0.8xx384pi^2` | ||
`=3031.942…` | ||
`=3032\ text{N (to 0 d.p.)}` |
An object is moving in simple harmonic motion along the `x`-axis. The acceleration of the object is given by `overset¨x = – 4 (x - 3)` where `x` is its displacement from the origin, measured in metres, after `t` seconds.
Initially, the object is 5.5 metres to the right of the origin and moving towards the origin. The object has a speed of 8 m s`\ ^(-1)` as it passes through the origin.
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i. `overset¨x = -4 (x – 3)`
`d/(dx)(1/2 v^2)` | `= -4x + 12` |
`1/2 v^2` | `= -2x^2 + 12x + c` |
`v = 8\ \ text(when)\ \ x = 0:`
`1/2 xx 8^2 = c \ => \ c = 32`
`1/2 v^2 = -2x^2 + 12x + 32`
`text(Find)\ \ x\ \ text(when)\ \ v = 0:`
`-2x^2 + 12x + 32` | `= 0` |
`x^2 – 6x – 16` | `= 0` |
`(x – 8)(x + 2)` | `= 0` |
`:.\ text(Particle oscillates between)\ \ x = -2\ \ text(and)\ \ x = 8`
ii. `overset¨x = -4 (x – 3) \ => \ n = 2`
`text(Amplitude = 5, Centre of motion at)\ \ x = 3`
`x = 5 cos(2t + alpha) + 3`
`text(When)\ \ t = 0, x = 5.5:`
`5.5` | `= 5cos alpha + 3` |
`cos alpha` | `= 1/2` |
`alpha` | `= pi/3` |
`:. x = 5 cos(2t + pi/3) + 3`
`text(Find)\ \ t\ \ text(when)\ \ x= 0:`
`cos(2t + pi/3)` | `= -3/5` |
`2t + pi/3` | `= 2.214…` |
`t` | `= 1/2(2.214… – pi/3)` |
`= 0.58\ text(seconds)\ \ text{(2 d.p.)}` |
A particle undergoing simple harmonic motion has a maximum acceleration of 6 m/s2 and a maximum velocity of 4 m/s.
What is the period of the motion?
`D`
`ddotx = -n^2 x`
`text{Find} \ n :`
`ddotx_text{max} = 6 \ \ text{occurs when} \ \ x = – a`
`6 = n^2 a\ …\ (1)`
`v^2 = n^2 (a^2 – x^2)`
`v_text{max} = 4 \ \ text{occurs when} \ \ x = 0`
`4^2` | `= n^2 (a^2-0)` | |
`16` | `= n^2 a^2` | |
`4` | `= n a\ …\ (2)` |
`text{Substitute} \ \ na = 4 \ \ text{from (2) into (1):}`
`6` | `= 4n` |
`n` | `= frac{3}{2}` |
`therefore \ text{Period} = frac{2pi}{n} = frac{4pi}{3}`
`=> \ D`
A particle starts from rest, 2 metres to the right of the origin, and moves along the `x`-axis in simple harmonic motion with a period of 2 seconds.
Which equation could represent the motion of the particle?
A. `x = 2cos pi t`
B. `x = 2 cos 2t`
C. `x = 2 + 2 sin pi t`
D. `x = 2 + 2 sin 2t`
`A`
`text(Period)` | `= 2` |
`(2 pi)/n` | `= 2` |
`n` | `= pi` |
`:.\ text(Eliminate B and D)`
`text(Particle starts at rest,)`
`(dx)/(dt) = 0\ \ text(when)\ \ t = 0`
`text(Consider A:)`
`(dx)/(dt) = -2pi sin (pi xx 0) = 0`
`text(Consider C:)`
`(dx)/(dt) = 2 pi cos (pi xx 0) = 2pi`
`=> A`
A particle is moving along the `x`-axis in simple harmonic motion centred at the origin.
When `x = 2` the velocity of the particle is 4.
When `x = 5` the velocity of the particle is 3.
Find the period of the motion. (3 marks)
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`2sqrt3pi`
`text(S)text(ince motion centred at origin,)`
`{:d/(dx):}^(1/2 v^2)` | `= −n^2x` |
`1/2 v^2` | `= int −n^2x\ dx` |
`= −1/2n^2x^2 + c` | |
`v^2` | `= c – n^2x^2` |
`text(When)\ v = 4, x = 2`
`16 = c – 4n^2\ …\ (1)`
`text(When)\ v = 3, x = 5`
`9 = c – 25n^2\ …\ (2)`
`text(Subtract)\ (1) – (2)`
`7` | `= 21n^2` |
`n^2` | `= 1/3` |
`n` | `= 1/sqrt3` |
`:.\ text(Period)` | `= (2pi)/n` |
`= 2sqrt3pi` |
A particle moves in a straight line. Its displacement, `x` metres, after `t` seconds is given by
`x = sqrt3\ sin\ 2t − cos\ 2t + 3`.
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i. `text(Show)\ \ ddot x = -4(x − 3)`
`x` | `= sqrt3\ sin\ 2t − cos\ 2t + 3` |
`dot x` | `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t` |
`ddot x` | `= -4sqrt3\ sin\2t + 4\ cos\ 2t` |
`= -4(sqrt3\ sin\ 2t − cos\ 2t)` | |
`= -4(sqrt3\ sin\ 2t − cos\ 2t + 3 − 3)` | |
`= -4(x − 3)\ \ …text(as required)` |
ii. `text(Period)\ = (2pi)/n`
`n^2 = 4 ⇒ n = 2\ \ text{(part (i))}`
`:.\ text(Period)` | `= (2pi)/2` |
`= pi\ \ text(seconds)` |
iii. `text(Write)\ \ dot x = 2sqrt3\ cos\ 2t + 2\ sin\ 2t`
`text(in form)\ \ \ A\ cos\ (2t − α)`
`A(cos\ 2t\ cos\ α + sin\ 2t\ sin\ α)` | `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t` |
`cos\ 2t\ cos\ α + sin\ 2t\ sin\ α` | `= (2sqrt3)/A\ cos\ 2t + 2/A\ sin\ 2t` |
`⇒ cos\ α` | `= (2sqrt3)/A` |
`⇒ sin\ α` | `= 2/A` |
`((2sqrt3)/A)^2 + (2/A)^2` | `= 1` |
`(2sqrt3)^2 + 2^2` | `= A^2` |
`:.A` | `= sqrt16` |
`= 4` |
`:.cos\ α` | `= (2sqrt3)/4 = sqrt3/2` |
`α` | `= pi/6` |
`:. dot x = 4\ cos\ (2t − pi/6)`
(iv) `text(Find)\ \ t\ \ text(when)\ \ dot x = ±2`
`text(If)\ \ dot x = 2`
`4\ cos\ (2t − pi/6)` | `= 2` |
`cos\ (2t − pi/6)` | `= 1/2` |
`2t − pi/6` | `= pi/3, 2pi − pi/3` |
`2t` | `= pi/2, (11pi)/6` |
`t` | `= pi/4, (11pi)/12` |
`text(If)\ \ dot x = -2`
`cos\ (2t − pi/6)` | `= – 1/2` |
`2t − pi/6` | `= (2pi)/3, (4pi)/3` |
`2t` | `= (5pi)/6, (3pi)/2` |
`t` | `= (5pi)/12, (3pi)/4` |
`:.t = pi/4, (5pi)/12, (3pi)/4, (11pi)/12\ \ text(seconds.)`
A particle is undergoing simple harmonic motion on the `x`-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds.
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i. `text{Amplitude (A)} = 18`
`text(Period) = (2 pi)/n = 5`
`5n` | `= 2 pi` |
`n` | `= (2 pi)/5` |
`text(Using)\ \ x` | `= A cos n t` |
`x` | `= 18 cos ((2 pi)/5 t)` |
ii. `text(When)\ \ t= 0,\ \ \ x = 18`
`text(Find)\ \ t\ \ text(when)\ \ x = 9`
`9` | `= 18 cos ((2 pi)/5 t)` |
`cos ((2 pi)/5 t)` | `= 1/2` |
`(2 pi)/5 t` | `= pi/3` |
`t` | `= (5 pi)/(3 xx 2 pi)` |
`= 5/6\ \ text(seconds)` |
`:.\ text(It takes the particle)\ \ 5/6\ \ text(seconds to move from)`
`text(rest position and half way to equilibrium.)`
A particle moves in a straight line and its position at time `t` is given by
`x = 5 + sqrt 3 sin3t - cos 3t.`
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i. `x = 5 + sqrt 3 sin 3t – cos 3t`
`sqrt 3 sin 3t – cos 3t` | `= R sin (3t – alpha)` |
`= R sin 3t cos alpha – R cos 3t sin alpha` |
`=> R cos alpha = sqrt 3\ \ \ \ \ R sin alpha = 1`
`R^2 = sqrt 3^2 + 1^2 = 4`
`R = 2`
`=> 2 cos alpha` | `= sqrt3` |
`cos alpha` | `= (sqrt3)/2` |
`alpha` | `= pi/6` |
`:.\ 2 sin ( 3t – pi/6) = sqrt 3 sin 3t – cos 3t`
ii. `x` | `= 5 + sqrt 3 sin 3t – cos 3t` |
`= 5 + 2 sin (3t – pi/6)` |
`text(Amplitude) = 2\ text(units)`
`text(Centre of motion at)\ \ x = 5`
iii. `text(Solution 1)`
`x = 5 + 2 sin (3t – pi/6)`
`dot x = 6 cos (3t – pi/6)`
`text(Maximum speed occurs when)`
`cos (3t – pi/6)` | `= 1` |
`3t – pi/6` | `= 0` |
`3t` | `= pi/6` |
`t` | `= pi/18` |
`text(Solution 2)`
`text(Maximum speed occurs at the)`
`text(centre of motion,)\ \ x = 5`
`5 + 2 sin (3t – pi/6)` | `= 5` |
`2 sin (3t – pi/6)` | `= 0` |
`sin (3t – pi/6)` | `= 0` |
`3t – pi/6` | `= 0` |
`t` | `= pi/18` |
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` ms`\ ^(–1)`. The parabola below shows `v^2` as a function of `x`.
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What are the values of `a`, `c` and `n`? (3 marks)
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i. `text(Particle is at rest when)\ v^2 = 0`
`:. x = 3\ \ text(or)\ \ 7`
ii. `text(Maximum speed occurs when)`
`v^2` | `= 11` |
`v` | `= sqrt11\ text(m/s)` |
iii. `v^2 = n^2(a^2 − (x − c)^2)`
`text(Amplitude) = 2`
`:.a = 2`
`text(Centre of motion when)\ x = 5`
`:.c = 5`
`text(S)text(ince)\ \ v^2 = 11\ \ text(when)\ \ x = 5`
`11` | `= n^2(2^2 − (5 − 5)^2)` |
`= 4n^2` | |
`n^2` | `= 11/4` |
`:.n` | `= sqrt11/2` |
Two particles oscillate horizontally. The displacement of the first is given by `x = 3\ sin\ 4t` and the displacement of the second is given by `x = a\ sin\ nt`. In one oscillation, the second particle covers twice the distance of the first particle, but in half the time.
What are the values of `a` and `n`?
(A) `a = 1.5,\ \ n = 2`
(B) `a = 1.5, \ \ n = 8`
(C) `a = 6,\ \ n = 2`
(D) `a = 6, \ \ n = 8`
`D`
`x_1` | `= 3\ sin\ 4t` |
`x_2` | `= a\ sin\ nt` |
`x_2\ text(has twice the amplitude of)\ x_1`
`:. a = 2 xx 3 = 6`
`x_2\ text(has a period)\ (T)\ text(that is half of)\ x_1`
`T(x_1)` | `= (2 pi)/n = (2pi)/4` |
`:.T(x_2)` | `= 1/2 xx (2 pi)/4 = (2pi)/8` |
`:.n` | `= 8` |
`=> D`
A particle is moving in simple harmonic motion in a straight line. Its maximum speed is 2 ms–1 and its maximum acceleration is 6 ms–2.
Find the amplitude and the period of the motion. (3 marks)
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`A=2/3\ text(m)`
`T=(2pi)/3\ text(sec).`
`text(Equations of SHM)`
`x` | `= A sin nt` |
`dot x` | `= An cos nt` |
`ddot x` | `= – An^2 sin nt` |
`text(Given max speed)\ = 2\ text(ms)^(-1)`
`text(and)\ \ \ -1 <= cos nt <= 1`
`=> An = 2\ \ \ \ \ …\ (1)`
`text(Given)\ \ ddot x \ \ text{(max)} = 6`
`=>An^2 = 6\ \ \ \ \ …\ (2)`
`text(Substitute)\ \ n=2/A\ \ text{from (1) into (2)}`
`A * (2/A)^2` | `= 6` |
`4/A` | `= 6` |
`A` | `= 2/3` |
`text(Substitute)\ \ A = 2/3\ \ text(into)\ (1)`
`2/3 n` | `= 2` |
`n` | `= 3` |
`text(Period) = (2pi)/n = (2pi)/3\ text(sec)`
`:.\ text(Motion has an amplitude of)\ \ 2/3 text(m)`
`text(and a period of)\ \ (2pi)/3\ text(sec.)`
A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that `v^2 + 9x^2 = k`, where `k` is a constant.
Show that the particle moves in simple harmonic motion with period `(2pi)/3`. (2 marks)
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`text(Proof)\ \ text{(See Worked Solutions)}`
`v^2 + 9x^2` | `= k` |
`v^2` | `= k\ – 9x^2` |
`1/2 v^2` | `= 1/2k\ – 9/2 x^2` |
`text(For SHM,)\ \ ddot x = -n^2x`
`ddot x` | `= d/(dx) (1/2v^2)` |
`= -9x` | |
`= -3^2 x \ \ \ text(… as required)` |
`text(Period)\ (T)\ text(of SHM) = (2pi)/n`
`text(Here,)\ \ n=3`
`:.T= (2pi)/3\ \ \ text(… as required)`
A particle is moving in simple harmonic motion with displacement `x`. Its velocity `v` is given by
`v^2 = 16(9 − x^2)`.
What is the amplitude, `A`, and the period, `T`, of the motion?
(A) `A = 3\ \ \ text(and)\ \ \ T = pi/2`
(B) `A = 3\ \ \ text(and)\ \ \ T = pi/4`
(C) `A = 4\ \ \ text(and)\ \ \ T = pi/3`
(D) `A = 4\ \ \ text(and)\ \ \ T = (2pi)/3`
`A`
`v^2 = 16(9 – x^2)`
`text(Find amplitude and period of motion)`
`v^2` | `= n^2(A^2 – x^2)` |
`A^2` | `= 9` |
`:.\ A` | `=3,\ \ \ (A > 0)` |
`n^2` | `= 16` |
`n` | `=4,\ \ \ (n>0)` |
`:. T` | `= (2pi)/n` |
`= pi/2` |
`=> A`