A 2.45-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension

Question

A 2.45-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 21.0 N. What are the frequencies of the first three allowed modes of vibration? f_1 = 4.63 Hz f_2 = 9.26 Hz f_3 = 13.89 Hz If a node is observed at a point 0.490 m from one end, in what mode and with what frequency is it vibrating? (Select all that apply.) The frequency is the fifteenth state at 69.4 Hz the frequency is the fifteenth state at 13.9 Hz the frequency is the fifth state at 11.6 Hz the frequency is the fifth state at 23.1 Hz the frequency is the third state at 13.9 Hz The frequency is the forty-fifth state at 208.3 Hz

Explanation / Answer

Here ,

L = 2.45 m

T = 21 N

m = 0.1 Kg

for the string

a) frequency of fundamentral node , f1 = sqrt(T/(m/L))/(2L)

frequency of fundamentral node , f1 = sqrt(21/(0.1/2.45))/(2 * 2.45)

frequency of fundamentral node , f1 = 4.63 Hz

f2 = 2 * f1 = 9.26 Hz

f3 = 3 * f1 = 13.89 Hz

b)

for a node to be 0.490 m away

wavelength = 2 * 0.490

wavelength = 0.98 m

frequency of wave = sqrt(21/(0.1/2.45))/(0.98)

frequency of wave = 23.1 Hz

also , wavelength = 0.490 m

frequency of wave = sqrt(21/(0.1/2.45))/(0.490)

frequency of wave = 46.3 Hz

for wavelenghth = 0.490/1.5

frequency of wave = sqrt(21/(0.1/2.45))/(0.490/1.50)

frequency of wave = 69.45 Hz

if take for the 45th state

frequency = 208.3 Hz , and it also has a node at 0.490 m away

hence , the correct options are

the frequency is the fifteenth state at 69.4 Hz

the frequecy is fifth state at 23.1 Hz

the frequency is forty fifth state at 208.3 Hz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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