A string with both ends held fixed is vibrating in its third harmonic. The waves

Question

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 191 m/a and a frequency of 235 Hz. The amplitude of the standing wave at an antinode is 0.45 cm.
A) calculate the amplitude at point on the string a distance of 23 cm from the left-hand end of the string.
B) how much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) calculate the maximum tranverse velocity of the string at this point.
D) calculate the maximum transverse acceleration of the string at this point. A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 191 m/a and a frequency of 235 Hz. The amplitude of the standing wave at an antinode is 0.45 cm.
A) calculate the amplitude at point on the string a distance of 23 cm from the left-hand end of the string.
B) how much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) calculate the maximum tranverse velocity of the string at this point.
D) calculate the maximum transverse acceleration of the string at this point.
A) calculate the amplitude at point on the string a distance of 23 cm from the left-hand end of the string.
B) how much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) calculate the maximum tranverse velocity of the string at this point.
D) calculate the maximum transverse acceleration of the string at this point.

Explanation / Answer

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192m/s and a frequency of 215 Hz. The amplitude of the standing wave at an antinode is 0.450 cm.

(A)Calculate the amplitude at point on the string a distance of 22.0cm from the left-hand end of the string.

(B)How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?

(C)Calculate the maximum transverse velocity of the string at this point.

(D)Calculate the maximum transverse acceleration of the string at this point.

answer

Wavelength = Speed/frequency = 192/215 = 89.3 cm

Angular frequency(w) = 2?f = 2? x 215 = 430? rad/s

(a) Standing Wave equation on the string :

y(x,t) = A cos(wt) sin(kx)   

(A = amplitude at antinode, w = Angular frequency, k = Wavenumber)

Since the time instant at which the wave is being sampled has not been mentioned, therefore assuming that the observations are being made at t = 0, therefore the standing wave equation changes to :

y(x) = (0.45 cm) sin(kx) (k = wavenumber = 2?/?)

? k = 2?/? = 2?/(89.3) = 0.07036 cm-1

? At x = 22 cm, y(x=22cm) = 0.45 sin(0.07036 * 22)

? y(x=22cm) = 0.45 * 0.9997 = 0.4498 cm (Ans)

(b) At a particular point, the time taken to go from the largest upward displacement to the largest downward displacement = Time period/2

? Time taken = T/2 = 1/2f (f = frequency)

= 1/(2*215) = 2.325 x 10-3 s = 2.325 ms (Ans)

(c) Transverse velocity = ?y/?t (partial derivative)

? ?y/?t = -Aw sin(wt) sin(kx)

For any time instant, ?y/?t will be maximum (positive) when sin(wt) = -1

? (?y/?t)max = Aw sin(kx)

Now at x = 22cm,

(?y/?t)max = (0.45)(430?) sin(0.07036 * 22) = 607.716 cm/s = 6.077 m/s

(d) Transverse acceleration = ?2y/?t2 (partial derivative)

? ?y/?t = -Aw sin(wt) sin(kx)

? ?2y/?t2 = -Aw2 cos(wt) sin(kx)

For any time instant, ?2y/?t2 will be maximum (positive) when cos(wt) = -1

? (?2y/?t2)max = Aw2 sin(kx)

Now at x = 22cm,

(?2y/?t2max = (0.45)(430?)2 sin(0.07036 * 22) = 820.954 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.