8. A steel guitar string of mass m and length 1.8 m is subjected to 3200 N of te

Question

8. A steel guitar string of mass m and length 1.8 m is subjected to 3200 N of tension, and is fixed at both ends. a. If the string's fundamental frequency when plucked is 225 Hz, what is the mass of the string? b. Suppose the amplitude of the fundamental wave is 2.1 mm. What is the average power transmitted by the wave? Suppose the tension is changed, and then 4 antinodes are created when the string is driven by an oscillator at 327 What is the tension now? What length would a string of the same material need to have in order that its fundamental frequency be 3 times the frequency of the string in (a)? c. d.

Explanation / Answer

a. Using the relation fro fundamental frequency, F= v/2L { v= speed of sound in wire= (T/)^1/2 where = mass per unit length= m/1.8 and T= 3200N; so v= (3200*1.8/m)^1/2= (5760/m)^1/2; and L= 1.8m }

so, F= ((5760/m)^1/2) / (2*1.8) = 225

which gives m= 0.00878kg

So, mass of string = m= 0.00878kg

b.The power transmitted by a transverse wave is given by P= 0.5**w^2*A^2*v { given A= 0.00211m, = mass per unit length= 0.00878/1.8=0.00487, v =(T/)^1/2= (3200/0.00487)^1/2= 809.96m/s; w= angular frequency= 2*pi*225= 1413.7 rad/s }

So, P= 0.5 * 0.00487 * 1413.7^2 * 0.00211^2 * 809.96 = 17.55W

So, Average power transmitted by the wave= 17.55W

c. Now, frequency= 327Hz and number of antidotes= 4. This means that string now is in 4th harmonis. So, using the relation for frequency for 4th hermonic, F= 4*F1= 327 {where F1 is frist harmonic}

So, F1= 327/4= 81.75

But, F1 is also given by F1= v/2L

so, 81.75= v/2*1.8

so, v= 294.3m/s

So, using v= (T/)^1/2;

294.3= (T/0.00487)^1/2

Which gives T= 421.8N

So, now the tension in string is 421.8N

d. Here, F= 3*225= 675Hz and since material of wire is same so = 0.00487 .

So, v= (T/)^1/2= (3200/0.00487)^1/2= 809.96m/s

Using the relation F= v/2L {where v= (T/)^1/2= (3200/(0.00878/L))^1/2 = 60.37*L^1/2}

675= 809.9/ 2L

Or, L = 0.6m

So, length of string of the same material for a fundamental frequency of 675Hz is 0.6m

please upvote to appreciate

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

---

---

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