How is tension distributed over the harp?
437418
I am uncertain, but I believe that the strings on a harp would have to oscillate via a certain mode corresponding to A*sin(n*pi) where n determines the mode of oscillation.  I would have to spend a lot more time though to determine the tension distribution on an actual harp though.
483991
You can use different string instruments or you an even use different types of strings. Rope, nylon, thread and use different tensions with them.
1026493
A pedal/concert harp has 46 or 47 strings.  Fork and lever harps have anywhere from 19 to 40 strings.  Early forms of the harp only had about ten strings.  These are the normal numbers, but it depends on the harp.
148349
http://www.harmonicaclub.com/common/note_bending_for_harmonica.htm\n\nthis is a good link too. it will show you how your tongue should be positioned. you can also listen to the clips to see to check your pitch. good luck!
280266
Force exerted on a material can cause stress and strain.  If you apply hand "pressure" to a glob of J-ello it will cause "stress" and deformation "strain."  If you set a weight on a surface, the force of the weight will cause "compressive stress" as it compresses the molecules of the surface that must strain slightly to bear the added force.  If you suspend a heavy magnet from a steel surface it causes "tensile stress" from tension that causes slight strain in the surface to support it.\n\nIf you take say a one square inch rod and press it down against a surface with a force of 100 pounds, the stress in the rod will be 100 PSI (pounds per square inch) compression and the stress on the surface will also be 100 PSI compression.  However, the force within the surface will become better distributed away from the contact point and the local stress at any such point will be less than 100 PSI.  Therefore, the cause and effect are both measured in PSI but depend for value on the distribution of the stress.  Although stress is calculated or measured in PSI the area in question may be much larger or much smaller than a square inch (like driving a car 60 MPH regardless of how long the trip is?).  Hope that helps a little.
921019
First,\ncompute the components of force perpendicular to the boom due to the weight of the boom and the object:\nBp=cos(48)*588\nOp=cos(48)*1770\n\nthe tension in the support cable can be computed by summing the torques around the pivot point:\n\nTC+TBp+TOp=0\nTC=tension*.66*24\nTBp=-Bp*.5*24\nTOp=-Op*24\nTension=(Bp*.5+Op)/.66\n=2093N\n\nThe vertical component of the tension is\ncos(48)*2093\nThe horizontal component is \nsin(48)*2093\n=1555N\nwhich is equal to the horizontal reaction force in the floor\n\nSince we know the components of force on the floor parallel to the beam as\nBh=sin(48)*588\nOh=sin(48)*1770\n\nthen v^2+h^2=parallel^2\nv=sqrt((sin(48)*588+sin(48)*1770)^2-1555^2)\n=807N\n\n\nj
617850
The same way you blow air through a trumpet
711880
The Cos of the pressure angle is equal to the Base Diamater over the pitch diamater of either gear in the gear set.  So since;\n\nCos(φ)=Dbase/Dpitch \n\nTherefore\n\nφ=cos^-1(Dbase/Dpitch )
386178
Below.\nWhy?\nBecause at 100ft above water you only feel the air pressure there. While at 100ft below water, you feel both the air pressure ( strictly spreaking in physics ) plus the water pressure there.\n\nAlso, note that water is more dense than air. This means for each space, water exerts more pressure compared to air for that same displacement of space.\n\nFurther, pressure that we feel is created due to the displacement of water or air moleculs respectively by the object. So, neighbouring air/water moleculs try to "displace out" the object that caused the "imbalance" where originally there were only air/water moleculs before the placement of the object.
636649
Air pressure and density. Sort of like when you fly, your ears pop.\nAs you get higher, air pressure and density are lower, than on normal ground level. This less pressure caused stress to build up in the bag as the air in the bag has more density than the air surrounding it.
112214
If you simulate a geometrical structure in the form of an arch you will have the most evenly distributed stress.and the moment will be the same at all points.\nIf your question refers to how you tied a stell structure together ,the modern method is glue it to gether with a special glue designed for that purpose.
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
760755
it is a wind instrument without reeds.the holes control the length of the vibrating column and hence the frequency of the note.it corresponds to the western flute
833112
First, I'm assuming that your question is "what do i do to make sure the length and mass per unit length of the wire is constant AS tension changes?" (note change of 'and' to 'as') and the rest of the material is just for explanation.\nThe answer is, you can't keep M/L constant. If your setup actually uses two fixed supports, L will be constant, but tensioning the wire stretches it, removing some M from between the supports. You can measure the amount of stretch by making two benchmarks (maybe small pieces of masking tape) in the unstretched wire and measuring their distance as you change the tension. (Doing the appropriate scaling between the baseline distance and the length of the stretched wire.)\nFor the theory on wire tension and resulting dimension changes, search Poisson's ratio. Basically stretching a length L of wire by epsilon*L (epsilon is a small number) would result in a thinning of radius R of 0.5*epsilon*R if the material were incompressible (thus inexpandable). However all solids including steel expand under tension to some extent, so the Poisson ratio of 0.5 above is more typically 0.25 to 0.35 for metals. See the ref.
1078605
Air is blown across reeds.  Kind of like a backward clarinet, which may account for the ... unique... sound.
154867
The best tension is 1 or 2 above the max. lbs given to have the racquet strung.\nThe strings that do good for me are Ballistic Polymomo from Babolot. I break normal strings once a week, but with these it's about once a month.\nAll of this does the best for me.
861607
The frequency, f, of a vibrating string of m kg/m and subject to a tension of T Newtons is given by the relation\n\nf=sqrt(T/m)/2/L\n\nHence the frequency is inversely proportional to the length of the string, assuming the tension remains unchanged.\n\nThus the required length of the string is\n40*474/624=30.385 cm.