Smh (1 Viewer)

[survivor]

ok heres a question that just stumped me

low tide at the mouth of a river occurs at 4pm and high tide at 10pm. low tide is 4m and high tide is 12m. a ship needs 10m of water to pass safely through the mouth of the river. if the motion of the tides is take to be simple harmonic, what will be the first times during which the ship may pass??

what the...
thankyou

 

[ezzy85]

the tides are from 4pm to 10pm. so your period is 6 hours. sub this into T = 2pi/n to find n.

low tide is 4m and high tide is 12m so the origin will be the average of the 2. ie. 8m. Therefore your amplitude will be from 4m to 8m. ie a = 4. once youve got these values sub them into the general cos curve. since it starts with low tide, use a minus cos curve. so x = -4cos(t pi/3 + @)

sub in the initial values to get @ and then youll get an equation for the motion of the waves. sub in x=10 and find t.

ill see if i can scan my working because i cant explain it properly without a diagram. sorry if ive confused you but basically, find the general equation of the curve and sub in what you know.

 

[wogboy]

Let t be the time in hours after 4pm (e.g. when t=3 it's 7pm), and let y be the height of the water above the ground (in metres) at time t.

First find the period of the SHM motion of the water tide (the time taken for one complete cycle of the tide). Since it takes 6 hours for a low tide to become high, it will take another 6 hours for the tide to become low again, so the period T = 6+6 = 12 (hours).

Also the equilibrium height is the average of the max & min heights (average of 4 and 12), which happends to be 8 (metres). The amplitude of the tidal height is 12 - 8 = 4 (metres). Now we have enough information to make an equation:

The general equation is:

y = +-A cos(2*(pi/T)*t) + X
(+- means plus or minus, A = amplitude, T=period, X=equilibium height)

y = -4*cos(2*(pi/12)*t) + 8
(the negative sign is since the tide starts low at 4pm or t=0)

y = 8 - 4*cos([pi/6] * t)

Now we need to find out when y=10.

10 = 8 - 4*cos([pi/6] * t)

solve for t, and you'll get t=4

i.e. it will be safe for the ships to start sailing when t=4. Translate this into normal time and you'll get your final answer to be 8pm.

 

[Newbie]

i have a simple question

A particle is moving with SHM of amplitude 10 metres. Find how long it takes to travel 6 metres from its mean position if the period is 10 seconds

help!
i dont even know where to start

 

[ezzy85]

youve got x = asin(nt + @)

a = 10

T=10 so 2pi/n = 10
n = pi/5

sub in x=6 and find t

is that close?

 

[Newbie]

what about @?

 

[Newbie]

wait nvm
you got it ez buddy
stupid question
who the hell uses radians these days

 

[survivor]

sorry to be really dumb but wogboy
when you sub for t how do you get t=4
can you do the working for me i cant figure it out?
sorry

 

[Starlette]

You guys might think I'm really dumb but I cant get this question could someone pleeaasseee help me?

A weight hanging from the roof on an elastic string is moving in SHM. It takes 4 seconds to move from the bottom of its motion, 15 cm above the floor, to the top of it's motion 55cm above the floor.

Find it's speed at the centre of the motion
and find the maximum accelaration?

Any help is much appreaciated