Vibration from a 475 Hz tuning fork sets up standing waves on a string clamped a

Question

Vibration from a 475 Hz tuning fork sets up standing waves on a string clamped at both ends (one of which you may take as x=0). The travelling wave speed on this string is 380 m/s. The standing wave has four anti-nodes and an amplitude of 4.0 mm.

(a) What is the length of the string?

(b) Write an equation for the transverse displacement of the string as a function of position and time:

y(x,t) = (___mm) sin (___m^-1 x)cos(___s^-1 t)

(c) What is the wavelength of the travelling waves that form this standing wave?

Explanation / Answer

let x be the distance from the rim to the water level in meters. so 2x will be the distance the sound travels to exactly match the next vibration of the tuning fork 1/2 the period of vibration = 1 / 475 (1/2) = 1 / 475(2) 1 / 475(2) sec = 2x m / 343 m/s (343) / 475(2) = 2x x = (343) / (475)(4) = 0.18 m or 18 cm from the top

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

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