Physics questions [updated sat 5 april]

yeah sup? im not much of a physics gangsta, so that's why im hoping you guys could help me. hopefully i can follow the steps, cos whenever i get it wrong the values always change.. *crazy* : D but it's the only way to learn : D

Stuck with this question:

A steel rod is 2.894 cm in diameter at 23.00°C. A brass ring has an interior diameter of 2.887 cm at 23.00°C. At what common temperature (in Celsius) will the ring just slide onto the rod? Give your answer to four significant figures. The linear expansion coefficient of steel is 11.00 x 10^-6 /C°. The linear expansion coefficient of brass is 19.00 x 10^-6 /C°.

edited

edited

click on to edit or quote-reply post.

thanks! im too used to vb

Q1) We start at the amplitude when t=0
Let's consider SHM from -d/2 to d/2

x(t) = Acos(wt), where A = d/2
x(0) = d/2
x(t) = d/4 = d/2 cos (wt)
cos (wt) = 1/2
wt = pi/3

but w = 2 * pi / T, where T = 40680s
so w = 2 * pi / 40680

wt = pi / 3 = 2 * pi * t / 40680
t = 40680 / (3*2) = 6780s = 113 mins

The rest I do later... A bit busy...

phew thanks.. i hit a wall at those questions, i managed to do some *removed those alrdy*

ps: what level of education are u at hey o.o!

Q2)

1st particle: x1(t) = Asin (wt)
2nd particle: x2(t) = Asin(wt + p), p = phase difference

when x1(t) = A/3, x1(t) = x2(t)
means x2(t) = A/3
sin (wt + p) = 1/3

also, x1(t) = A/3
sin(wt) = 1/3

since sin inverse 1/3 = 19.5 degrees or 160.5 degrees (less than 180)

p = 160.5 - 19.5 = 141 degrees.

I'm not very sure if the answer is correct... It is logical because if the phase difference is 180 degrees, they will pass each other in opposite direction at displacement = 0.

yeah thats correct hey ^^

k i suck at harmonic motion so i tried qn 7..

at equilibrium point the velocity is 80cm/s.. so the KE at that point( which is the max) is = 1/2 mv^2, so is 1/2 x 1.7kg x 0.8m x 0.8m = 0.544J.. so just find the point where 0.544J is on the vertical axis and draw horizontal line and can get x = 11 or 12cm? if im wrong pls tell me hor lol..

just feeling bored and nothing to do..

3) A sinusoidal wave of frequency 670 Hz has a speed of 340 m/s. (a) How far apart are two points that differ in phase by pi/5 rad? (b) What is the phase difference between two displacements at a certain point at times 1.6 ms apart?

I don't understand part (a)
How come got distance... Is it just distance in amplitude? Or is the horizontal distance taken into account too, i.e. finding the diagonal distance?

4) The linear density of a string is 1.3 × 10-4 kg/m. A transverse wave on the string is described by the equation
y = (0.015 m) sin[(2.8 m-1)x + (31 s-1)t]
What are (a) the wave speed and (b) the tension in the string?

Can define what is m (metres?), s and t (time?) Same for question 5.

Thanks

i think when it says distance, it means just the amplitude (y-axis) because a particle undergoing shm only goes up and down.

as for ur other question, im not sure, perhaps they are units?

here's a pic with the proper format

http://i105.photobucket.com/albums/m225/nickykeeng/album2/physics1.png

Originally posted by WoAiMeiMei:

phew thanks.. i hit a wall at those questions, i managed to do some *removed those alrdy*

 

ps: what level of education are u at hey o.o!

He is some super pro guy.

Originally posted by WoAiMeiMei:

i think when it says distance, it means just the amplitude (y-axis) because a particle undergoing shm only goes up and down.

 

as for ur other question, im not sure, perhaps they are units?

here's a pic with the proper format

http://i105.photobucket.com/albums/m225/nickykeeng/album2/physics1.png

I'm thinking it is a sine wave, so there might be horizontal distance too, which can be calculated easily by the time difference * the speed

I go jogging first, then come back attempt

Q3)
Damn, I think I really too long never touch physics till I have to google for answer... Sorry for making everyone confused earlier... Distance is meant by horizontal.

Anyway refer to this website on google book

The equation is given by:

in our case,

Φ  = π / 5

and

λ =  v / f = 340 / 670

we want to find x.

Substituting values into the equation,
x = 0.0507 m

Q4 and Q5 I no inspiration at the moment... do later...

Q6)
Two sound waves, from two different sources with the same frequency, 665 Hz, travel in the same direction at 335 m/s. The sources are in phase. What is the magnitude of the phase difference of the waves at a point that is 4.18 m from one source and 5.17 m from the other?

λ = 335/665 = 0.50376m

At 4.18m, this is 8.2976λ
At 5.17m, this is 10.2628λ

The difference is 1.9652λ

Using the same equation as in Q3,

phase difference = 2 * pi * 1.9652 = 3.9π  --> not confirmed... Sometimes  questions don't take greater than π, and if that's the case, your answer will be 0.1π.

my eagle is a freak in NUS hor...LOL.

oh ya, TS, wad level are u in now?

Originally posted by hiphop2009:

my eagle is a freak in NUS hor...LOL.

oh ya, TS, wad level are u in now?

From the questions, I guess A levels. SHM is my weakest portion in physics, so maybe I cannot really judge totally...

it's first year engineering advanced physics >.<

i will check the answers later, and substitute new values cos everytime i get it wrong, values change hey. wait ill eat first

with question 6,  how to you get this bit?

At 4.18m, this is 8.2976λ
At 5.17m, this is 10.2628λ

these are like the new values /sigh

Two sound waves, from two different sources with the same frequency, 672 Hz, travel in the same direction at 335 m/s. The sources are in phase. What is the magnitude of the phase difference of the waves at a point that is 7.23 m from one source and 4.32 m from the other?

try Q7, it looks easy but i dunno where to start hey

lemme try...

The linear density of a string is 1.3 × 10-4 kg/m. A transverse wave on the string is described by the equation
y = (0.015 m) sin[(2.8 m-1)x + (31 s-1)t]
What are (a) the wave speed and (b) the tension in the string?

2π/λ = 2.8, 2π/T = 31,

λ = 2.244 m, T = 0.2027 s

v = fλ = λ/T = 11.1 ms-1 = 11 ms-1 (2sf)

v = sqrt(Tension/linear density),

Tension = v^2 * linear density = 1.4 * 10^-3 N (2sf)

Originally posted by WoAiMeiMei:

with question 6,  how to you get this bit?

At 4.18m, this is 8.2976λ
At 5.17m, this is 10.2628λ

these are like the new values /sigh

Two sound waves, from two different sources with the same frequency, 672 Hz, travel in the same direction at 335 m/s. The sources are in phase. What is the magnitude of the phase difference of the waves at a point that is 7.23 m from one source and 4.32 m from the other?

 

try Q7, it looks easy but i dunno where to start hey

new values are:

λ = 335/672 = 0.4985m

7.32m / 0.4985m = 14.684λ
4.32m / 0.4985m = 8.666λ

Originally posted by angelfairy:

lemme try...

 

The linear density of a string is 1.3 × 10-4 kg/m. A transverse wave on the string is described by the equation
y = (0.015 m) sin[(2.8 m-1)x + (31 s-1)t]
What are (a) the wave speed and (b) the tension in the string?

2π/λ = 2.8, 2π/T = 31,

 

λ = 2.244 m, T = 0.2027 s

 

v = fλ = λ/T = 11.1 ms-1 = 11 ms-1 (2sf)

 

v = sqrt(Tension/linear density),

 

Tension = v^2 * linear density = 1.4 * 10^-3 N (2sf)

 

 

I'm not very clear about the formulas... But suppose they are correct, Q5 can be done in a very similar way.

yes! ur method works fine =]

man.. every week i have like to do 20 of these damned questions =[

some of them are like o.o?

5) A string oscillates according to the equation
y´ = (0.734 cm) sin[(pi/3.0 cm^-1)x] cos[(22.7 pi s^-1)t].
What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position x = 2.27 cm when t = 1.19 s?
Give your answers in centimeter-based units.

Background of question: From the equation and the questions, I think this is the equation of a standing wave. See wiki for more info.

a) Since x and t are unrelated variables (or so I think),
amplitude = 0.734cm

b) This one I base my equations on Q4 as kindly answered by angelfairy.
2π/λ = π/3, 2π/T = 22.7π
λ=6cm, and T = 2/22.7 = 0.088s

thus, f = 1/T = 11.35 s-1

speed = fλ = 68.1 cm/s

c) Nodes are where y is always 0. This happens at the point where the sin component is 0 since it is dependent on x, the location of the wave.

so, [ π/3.0 cm^(-1) ] x = 0 or π (2 ajacent nodes. You can try π and 2π also, just more tedious)

x = 0 or x = 3
so distance = 3-0 = 3cm

d) dy/dt = (0.734 cm) sin[(pi/3.0 cm^-1)x] ( -sin[(22.7 pi s^-1)t] )
substitute values x = 2.27 cm when t = 1.19 s

should be work in radians I guess

dy/dt = 0.0207 cm/s

I not very sure for c and d