Figure 16.8). The length of each string between its two fixed ends is 0.628 m, a

Question

Figure 16.8). The length of each string between its two fixed ends is 0.628 m, and the mass is 0.208 g for the highest pitched E string and 3.32 g for the lowest pitched E string. Each string is under a tension of 226 N. Find the speeds of the waves on the two strings. AM and FM radio waves are transverse waves consisting of electric and magnetic disturbances traveling at a speed of 3.00 times 10^8 m/s. A station broadcasts an AM radio wave whose frequency is 1230 times 10^3 Hz (1230 kHz on the dial) and an FM radio wave whose frequency is 91.9 times 10^6 Hz (91.9 MHz on the dial). Find the distance between adjacent crests in each wave.

Explanation / Answer

19. for highest pitched:

linear mass density = 0.208 x 10^-3 kg / 0.628 m

= 3.31 x 10^-4 kg/m

v = sqrt[ T / linear mass density]

v = sqrt[ 226 / (3.31 x 10^-4)]

v = 826 m/s

For lowest pitched:

linear mass density = (3.32x 10^-3) / (0.628) = 5.29 x 10^-3 kg / m

v = sqrt[ 226 / (5.29 x 10^-3)]

v = 206.8 m/s

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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