A transverse wave is traveling on a string. The displacement y of a particle from its...
Q1) The following equation describes a transverse wave on a string: The displacement y of a particle from its equilibrium position is given by: y 0.02 1sin (2.0x-2. 5t) Note: the phase angle is in radians, t is in seconds, x and y are in meters. Determine: a) b) c) d) The amplitude of the wave The frequency of the wave The wavelength of the wave The speed of the wave
A sinusoidal transverse wave is traveling along a string in the negative direction of an x axis. The figure below shows a plot of the displacement as a function of position at time t = 0. The x axis is marked in increments of 10 cm and the y axis is marked in increments of 2 cm. The string tension is 3.1 N, and its linear density is 34 g/m. (a) Find the amplitude. m (b) Find the wavelength. m...
The displacement of a transverse traveling wave on a string under tension is described by: D(x, t) = (2.0 cm) .sin((12.57 rad/m)x + (638 rad/s)t + /2] The linear density of the string is 5.00 g/m. 1. What is the tension in the string? 2. What is the maximal speed of a point on the string? String 2 3. The original string (String 1) is tied to a second string with String 1 a linear density of 12 g/m, as...
The displacement of a transverse traveling wave on a string under tension is described by: D(x, t) = (2.0 cm) sin((12.57 rad/m)x+ (638 rad/s)t + T/2] The linear density of the string is 5.00 g/m. 1. What is the tension in the string? 2. What is the maximal speed of a point on the string? String 2 3. The original string (String 1) is tied to a second string with String 1 a linear density of 12 g/m, as shown...
The equation of a transverse wave traveling in a string is given by y = A sin(kx-at). The tension in the string is 18.0 N, A What is the wave speed? 1 mm, k = 26 rad/m, ω 745 rad/s Submit Answer Tries 0/99 What is the linear density of the string. Submit Answer Tries 0/99
The equation of a transverse wave traveling along a very long string is y = 3.96 sin(0.0444πx+ 7.89πt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 1.05 cm when t = 0.843 s?
The equation of a transverse wave traveling along a very long string is y = 6.28 sin(0.0223πx+ 3.63πt), where x and yare expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 4.95 cm when t = 0.876 s?
The equation of a transverse wave traveling along a string is y = 0.419 sin(0.265x - 18.90), in which x and y are in meters and t is in seconds. (a) What is the displacement y at x = 6.36 m, t = 0.582 s? (Hint: Displacement is a vector quantity. Pay attention to the sign.) -.0442 m(b) Choose an equation of a wave that, when added to the given one, would produce standing waves on the string. O V'(x,t)...
Parts E-H please A sinusoidal transverse wave is traveling along a string in the negative direction of an x axis. The figure shows a plot of the displacement as a function of position at time t 0; the y intercept is 4.0 cm. The string tension is 2.1 N, and its linear density is 21 g/m. Find the (a) amplitude, (b) wavelength, (c) wave speed, and (d) period of the wave. (e) Find the maximum transverse speed of a particle...
The equation of a transverse wave traveling along a very long string is given by y = 6.1 sin(0.018πx + 3.1πt), where x and y are expressed in centimeters and t is in seconds. Determine the following values. (a) the amplitude cm (b) the wavelength cm (c) the frequency Hz (d) the speed cm/s (e) the direction of propagation of the wave +x−x +y−y (f) the maximum transverse speed of a particle in the string cm/s (g) the transverse displacement at...