Simple harmonic motion

B

Blake Andrews

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Feb 12, 2020
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(a) Show that a particle is moving according to SHM if its displacement is given by x = 5cos 3t + 2sin 3t, where x is in metres and t is in seconds.
(b) Find the maximum speed.

(a) x = 5cos 3t + 2sin 3t

v = -15sin 3t + 6cos 3t

a = -45cos 3t - 18sin 3t
= -9(5cos 3t + 2sin 3t)
= -9x

Acceleration is proportional to displacement so the particle is moving according to SHM.

(b) I'm having trouble with this part. I was thinking of finding the equation for v in terms of x and then use v is max when x = 0:

1/2v2 = ∫-9x dx
= -9x2/2 + C

However i don't think i have enough information to find C.

Maybe i could find t when x = 0 and then find v but i'm not sure how to solve for t when 5cos 3t + 2sin 3t = 0
 
Blake Andrews said:
(a) Show that a particle is moving according to SHM if its displacement is given by x = 5cos 3t + 2sin 3t, where x is in metres and t is in seconds.
(b) Find the maximum speed.

(a) x = 5cos 3t + 2sin 3t

v = -15sin 3t + 6cos 3t

a = -45cos 3t - 18sin 3t
= -9(5cos 3t + 2sin 3t)
= -9x

Acceleration is proportional to displacement so the particle is moving according to SHM.

(b) I'm having trouble with this part. I was thinking of finding the equation for v in terms of x and then use v is max when x = 0:

1/2v2 = ∫-9x dx
= -9x2/2 + C

However i don't think i have enough information to find C.

Maybe i could find t when x = 0 and then find v but i'm not sure how to solve for t when 5cos 3t + 2sin 3t = 0
Click to expand...
How do you usually find a function's max and min points?
 
I would let dy/dx equal 0 to find any stationary points and determine their nature.

So if i want to find the stationary points of v = -15sin 3t + 6cos 3t i would let dv/dt = 0.

-45cos 3t -18sin 3t = 0 which i'm unsure how to solve.

I had a look at v = -15sin 3t + 6cos 3t on a graphing calculator and the amplitude of v is the answer i'm looking for but i'm unsure how to calculate it.
 
Use trig identities. That is why they beat them into your head in pre-calculus.

[MATH]\text {Let } u = 3t \implies t = u/3 \text { and } - 45cos(u) - 18sin(u) = 0 \implies [/MATH]
[MATH]-18sin(u) = 45cos(u) \implies sin(u) = -\dfrac{45cos(u)}{18} \implies tan(u) = - 2.5.[/MATH]
Now what?

EDIT: Be careful.
 
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JeffM said:
Use trig identities. That is why they beat them into your head in pre-calculus.

[MATH]\text {Let } u = 3t \implies t = u/3 \text { and } - 45cos(u) - 18sin(u) = 0 \implies [/MATH]
[MATH]-18sin(u) = 45cos(u) \implies sin(u) = -\dfrac{45cos(u)}{18} \implies tan(u) = - 2.5.[/MATH]
Now what?

EDIT: Be careful.
Click to expand...
Be vewy vewy careful - that waskelly negative sign will throw the solution out of first quadrant
 
I still struggle finding these steps sometimes, i really need to memorise the trig identities.

tan (3t) = -2.5

3t = -68.2

Sub this into v = -15sin 3t + 6cos 3t

= 16.2 ms-1

Was the warning about the negative sign to make sure i didn't sub +68.2 into v?
 
Blake Andrews said:
I still struggle finding these steps sometimes, i really need to memorise the trig identities.

tan (3t) = -2.5

3t = -68.2

Sub this into v = -15sin 3t + 6cos 3t

= 16.2 ms-1

Was the warning about the negative sign to make sure i didn't sub +68.2 into v?
Click to expand...
tan (3t) = -2.5

3t = -68.2 or (-68.2+180 = ) 111.8
 
The "be careful" was to make sure that you address all the little fussy bits. Are you dealing with radians or degrees? Are there multiple solutions? What is the second derivative? Remember that the the arctan function has a limited domain?
 
[MATH]x = 5\cos(3t) + 2\sin(3t)[/MATH]
[MATH]C\sin(3t) = 5[/MATH], [MATH]C\cos(3t) =2[/MATH]
[MATH]C^2\sin^2(3t) = 25[/MATH], [MATH]C^2\cos^2(3t) = 4[/MATH]
[MATH]C^2[\sin^2(3t) + \cos^2(3t)] = 29 \implies C = \sqrt{29}[/MATH]
[MATH]x = C\sin(3t)\cos(3t) + C\cos(3t)\sin(3t)[/MATH]
[MATH]x = \sqrt{29} \cdot 2\sin(3t)\cos(3t) = \sqrt{29}\sin(6t)[/MATH],
note [math]\omega = 6 \implies T= \dfrac{\pi}{3}[/math]
[MATH]v = 6\sqrt{29} \cos(6t) \implies |v_{max}| = 6\sqrt{29}[/MATH]
[MATH]a = -36 \sqrt{29} \sin(6t) = -\omega^2 x[/MATH] , indicating acceleration is opposite in direction and directly proportional to displacement ... the key characteristic of SHM.
 
skeeter said:
[MATH]x = 5\cos(3t) + 2\sin(3t)[/MATH]
[MATH]C\sin(3t) = 5[/MATH], [MATH]C\cos(3t) =2[/MATH]
[MATH]C^2\sin^2(3t) = 25[/MATH], [MATH]C^2\cos^2(3t) = 4[/MATH]
[MATH]C^2[\sin^2(3t) + \cos^2(3t)] = 29 \implies C = \sqrt{29}[/MATH]
[MATH]x = C\sin(3t)\cos(3t) + C\cos(3t)\sin(3t)[/MATH]
[MATH]x = \sqrt{29} \cdot 2\sin(3t)\cos(3t) = \sqrt{29}\sin(6t)[/MATH],
note [math]\omega = 6 \implies T= \dfrac{\pi}{3}[/math]
[MATH]v = 6\sqrt{29} \cos(6t) \implies |v_{max}| = 6\sqrt{29}[/MATH]
[MATH]a = -36 \sqrt{29} \sin(6t) = -\omega^2 x[/MATH] , indicating acceleration is opposite in direction and directly proportional to displacement ... the key characteristic of SHM.
Click to expand...

I was trying to follow your working but couldn't see where C sin 3t = 5 and C cos 3t = 2 come from.

I found another example using:

a sin x + b cos x = C sin (x + a)

2 sin 3t + 5 cos 3t = C sin (3t + a)

= C sin 3t cos a + C cos 3t sin a

C cos a = 2
C sin a = 5

C2 cos2 a = 4 (1)
C2 sin2 a = 25 (2)

(1) + (2):

C2(sin2 a + cos2 a) = 29

C = sqrt29

x = sqrt29 sin (3t + a)

v = 3sqrt29 cos (3t + a)

Amplitude of v is 3sqrt29 = 16.2 ms-1
 
yes, I screwed up ... Saturday night, don’t drink and derive.

[MATH]x = \sqrt{29} \cos(3t-\alpha)[/MATH], [MATH]\alpha = \arctan(2/5)[/MATH]
 
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