Is wave speed same as speed of any part of string for transverse wave?
No. The actual speed at which any part of the string vibrates need not necessarily be the same as the wave speed. Average Speed at which any part of the string vibrates = 4 * Amplitude * frequency. Speed of wave propagation = frequence * wavelength. If amplitude = 4 * wavelength, then the two speeds would be equal, however this would just be a coincidence and not required to be true in all cases.
V=√(T/Mu).Where T is the tension ,Mu is mass per unit length. In this case Mu being the mass per unit length,here the length of string is constant and doubling its diameter could increase its mass duadraupled (4times) ,making the Mu 4times.So V becomes √(1/4)=(1/2) the original.
V= √T/u.Thus if T becomes 4 times v becomes twice
[math]F=ma[/math][math]T=\mu L \times \frac {v}{t}[/math], L=length of string, v is velocity of wave.[math]\frac {T}{\mu}=\frac {Lv}{t}[/math][math]\frac {L}{t}=v[/math][math]\frac {T}{\mu}=v^2[/math][math]v=\sqrt {\frac {T}{\mu}}[/math]
Maximum Transverse Speed Of a Particle on a Wave Physics?
The displacement in the transverse direction at time t and horizontal location x is given by the equation in the problem: y(x,t) = (7 cm)*sin((0.017 cm^-1)*x + (4.4 sec^-1)*t) The transverse *velocity* at time t and horizontal location x is given by the derivative of the above equation with respect to time: v_y(x,t) = d/dt (y(x,y)) = (30.8 cm/sec)*cos*((0.017 cm^-1)*x + (4.4 sec^-1)*t) The cosine function oscillates between +1 and -1, so the maximum speed occurs when the cos function is at it's extrema, and the magnitude of that speed is simply given by the factor in front of the cos function. The maximum |v_y| = 30.8 cm/sec. This velocity is reached when (0.017 cm^-1)*x + (4.4 sec^-1)*t = n*pi, where n = 0, 1, 2, ...... This are the locations where the displacement is at a minimum (because the sin(z) = 0 when |cos(z)| = 1.)
How do I find maximum transverse speed of a particle on a wave. PHYSICS?
the displacement of the particle along the y axis at any time (t) at a given (x) along x-axis y = 7.0 sin (0.017 x + 4.4 t) >>>>{{= a sin (k x + w t)}} speed of particle v(y) = dy/dt >>> differentiate wrt (t) v(y) = 7.0 cos (0.017 x + 4.4 t) *[4.4] v(y) = 30.8 cos (0.017 x + 4.4 t) >>>cm/sec -----(1) v(y) = 30.8 [1 - (y/7)^2]^1/2 ---(1) --------------------------------------... for this speed to be maximum cos (0.017 x + 4.4 t) = +-1 or sin (0.017 x + 4.4 t) = 0 >>>> y =0 >>>>NODES >>(1) --------------------------------------... for this speed to be maximum y=0 from (2) ----------------------------------- V(max) = 30.8 cm/sec v(min) at y = 7 >>> sin [(0.017 x + 4.4 t)] = +-1 Antinodes v(min) =0 tangent >>> measuring velocity of particle becomes parallel to x-axis
Transverse waves propagating down a string under tension travel at a speed given by v = sqrt(T/d), where T is the string tension and d is the string’s linear density, or d = m/L (mass per unit length). Thus doubling the diameter quadruples the linear density since mass is proportional to cross sectional area and area goes as (diameter)^2; with v proportional to the square root of the inverse of d, the speed of the waves will therefore be cut in half by doubling the diameter.
Calculate the theoretical speed of transverse waves on the same string?
Speed is = Square root of (tension in string/ mass per unit length)