2 waves are travelling in the same direction along a stretched string and are 90 degrees out of phase with amplitude 10cm each. Whats the amplitude of the resultant wave? How to go abt calculating it?
201302
the maximum amplitude is 10 times square root of 2. if the first wave is a sine wave, then any wave shifted by 90 degrees is the same as a cosine wave. so you have to find the maximum value of sin x + cos x. to find the maximum value of a function, set its derivative equal to zero, so you get cos x - sin x = 0. this is true when x = 45 degrees and sin x = cos x = 1 / square root of 2. when you add the two, you get square root of 2, which you multiply by the initial amplitude.
278559
FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
543823
The sound intensity due to a point source radiating equally in all directions (spherically symmetric case) is proportional to 1/r^2, where r is the distance from the source of the sound:\n\nI = k/r^2, where k is a constant of proportionality.\n\nIf we know the intensity (I1) at one distance (r1), then the intensity (I2) at a second distance (r2) is given by:\n\nI2 = I1*(r1/r2)^2\n\nIn Q1, we are told that the sound intensity at 1 m is 10^-5 W/m^2, and we want to know the intensity at 5 m.  Plugging these values into the above equation, we have:\n\nI2 = 10^-5 W/m^2 * (1m/5m)^2 = (10^-5 W/m^2)/25 = 4*10^-7 W/m^2\n\nThe intensity of a sound wave is proportional to the square of the amplitude of the displacement of a particle in the medium through which the sound is propagating:\n\nI = m*x^2,\n\nwhere m is a constant of proportionality, and x is the amplitude of the displacement.  Plugging this into the equation above that relates intensity and distance, we have that:\n\n(x2)^2 = (x1)^2 * (r1/r2)^2\nx2 = x1*r1/r2\n\nThe amplitude of the displacement is inversely proportional to 1/r.\n\nFor Q1 we are told that x1  = 70 microns at 1 meter, and are asked what x2 is at 5 meters.  Plugging these values into the above equation, we have:\n\nx2 = 70 microns * 1/5 = 14 microns\n\nFor Q2, we are told that x1 = 10^-7 m at r1 = 3 m, and are asked what the amplitude is at 5 m.  Again plugging these into the above equation, we have:\n\nX2 = 10^-7 m * (3/5) = 6*10^-8 m
21870
To make them interfere destructively, they will have to be half a wavelength out of sync.\n\n\nSpeed = 340 m/s\nThat means in one second, the soundwave travels 340 meters.\n\nFrequency = 1960 Hz\nThat means in one second, 1960 cycles are produced.\n\n\nCombine those two facts, and you can calculate the wavelength.\n\nIn one second, there will be 1960 cycles in 340 meters. So wavelength = 340 meters / 1960 cycles.\n\nWavelength = 0.1734m\n\nHalf of wavelength = 0.0867m
344207
Well this is just like sound measurement. First of all you can calculate the amount of energy released from the eq. That would be equal to 2000000*(pi*48000^2)J/s = 1.45 x 10^16 J/s\ntherefore at a distance of 1km, it would be 1.45 x 10^16 J / (pi*1000^2)sm^2 = 4.61 x 10^9 J/sm^2. \n\nRate of energy passing through 5km^2 would be (4.61 x 10^9)x(5x1000)^2 = 1.152 x 10^17 Watts = 115200 Tera Watts
400564
This is a classic problem in 1-D wave propagation.\nWhen the medium is limp (i.e. you can ignore its stiffness) then the speed of propagation of a transverse wave is sqrt(T/mu) where mu is the mass per unit length and T is the tension.\n\nT is the same for all segments. So in the first section the wave propagates at sqrt(T/mu), in the second segment at 1/sqrt(2) as fast, in the third segment at 1/2 as fast.\nThe time to traverse each segment is the length of the segment divided by the speed, and all segments are L/3 long.\nTherefore the time to reach the wall is approximately\n1.47 L * sqrt(mu/T)\nif I have done the algebra correctly.
533797
you would hear the resultant of the two frequencies both in frequencie and amplitude.
443109
First figure out whether to use sin or cos, I decided on cos because we look for the max and min.\n\nNext find the amplitude which is 6. a = amplitude/2 so a*cos(x).\n\nThird find the midpoint of the curve which is (max+min)/2 = m. a*cos(x) + m.\n\nThe fourth step is to get the max or min at the x cord. that you want. With cos you subtract max from x. a*cos(x-max) + m.\n\nFinally you find your wave length, which is simply (max-min)/pi = wl for cos. a*cos((x-max)/wl) + m\n\nso you go from (-2,1) to (10,7)\n\na = 3\nm = 4\nmax = 10\nwl = 12/pi to give us\n\n3*cos ((x-10)/(12/pi)) + 4
446739
The phase difference at the point under consideration is an odd multiple of pi, as the vibrations annul each other.\n\nThe phase difference is given by the equation 2pi (R1 - R2)/ L where L is the wavelength of the wave, R1 and R2 are the distances of the point from the sources of the sound.\n\nC = N L where C is the velocity of the wave, N is the frequency and L is the wave length.\n\nTherefore, 2pi (R1 - R2) N = (2 k +1) pi C, where k = 0. 1, 2, 3, etc.\n\nTherefore 2 (R1 - R2) 500 = (2 k +1)350.\n\n(R1 - R2) = (2k +1) x 0.35 m\n\nThe distances will be 5 plus or minus (2k +1) x 0.35 m where k = 0, 1, 2, 3, etc.\n\n\nIf k = 0, the distance will be 5.35 or 4.65 m
426621
v = wave length * f.\n\nSo, length = v / f =   1,7m
387170
If a weight is hanging from a V formed by two ropes, the angle of the V cannot be 180 deg, unless the weight is 0.\n\nThis is because the Tensions in the two ropes must compose to a force equal and opposite to the weight, which is acting vertically down.\n\nIf the angle is 0, then each tension is half the weight = 500/2 = 250 N\n\nFor any other angle, we need to know the relative lengths of the ropes.  If they are equal, the tensions are equal and\n2Tcos(a/2) = weight.\nwhere a is the "V" angle.
549967
You can easily calculate the speed of the transverse wave with the given information for distance travelled and time elapsed, with the formula s=vat where s is the distance va is the magnitude of velocity and t is the time.  Rearange that to get your speed.\n\nTHe speed in strings is dependent on the tension and the linear density of the string.  The formula is Vstring= square root (Tstring/linear density of the string)  Rearange this formula to calculate your tension.  Note:  the linear desnity can be calculated with the given information of the mass of the string and its distance...I think.
351206
It's two.  Energy goes as the square of amplitude.
213052
Try these answers:\n\nFor the first one, shortest is 0.0086 m, longest is 8.6 m.\nFor the second one, it should be 2.95 Hz
115057
Answering...\n\nBasic Equation:\n\nSine:.. y = a sin bx\nCosine: y = a cos bx\nwhere:\namplitude = absolute value of a (always positive)\nPeriod = 360 degrees/absolute value of b or 2pi/abs b\n\nEX: y = sin 4x\namplitude = abs a = 1\nPeriod = 360 deg./abs. b = 360/4 = 90 deg.\n\nEXPLANATION:\n\nSine Function\n1. The graph looks like a regular sea wave.\n2. For positive function, along the y-axis, the graph starts at the origin and then it goes up like a wave until it reaches the peak (called crest).\n3. Then it goes down until it reaches its lowest part (called trough).\n4. Then the cycle is repeated.\n5. For a negative sine function, from the origin, the graph goes down to its trough and then up again.\n\nCosine Function:\n1. The graph also looks like a regular sea wave.\n2. For a positive function, the graph (along the y-axis) starts at the its crest.\n3. Then it goes down until it reaches its trough.\n4. Then it goes up again until it reaches its crest again.\n5. The cycle is repeated.\n6. If the function is negative, along the y-axis, the graph starts at its trough first. Then it goes up.
441843
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°
370634
You create two waves a second -> frequency = 2 Hz\n\nperiod = 1/f = 1/2 = 0.5 s\n\nThe wave travels 0.18m per 0.5 s, which is .36 m/s