Exercise 1 – Wave Speed and Linear Density
Two strings are stretched between two supports 2.00 m apart, both under a tension of 600 N. String
1 has a linear density µ■ = 0.0025 kg/m, and String 2 has µ■ = 0.0035 kg/m. A transverse pulse is
sent simultaneously from each end toward the center.
Question: How long will it take before the two pulses meet?
Diagram: Two strings between fixed supports (v■ and v■ labeled).
Show your work here:
Exercise 2 – Standing Waves on a String
A string vibrator is attached to a string passing over a pulley and supporting a hanging mass. The
length of the vibrating part of the string is L = 1.00 m and the linear density is µ = 0.006 kg/m. The
hanging mass is m = 2.00 kg.
(a) Find the wavelength and frequency for the n = 6 mode.
(b) If this vibration produces a sound wave in air (v■ = 343 m/s), what is the wavelength of the
sound wave?
Diagram: String vibrator – pulley – hanging mass system (T = mg).
Exercise 1 – Solution
Wave speed on a string: v = √(F/µ).
v■ = √(600/0.0025) = 490 m/s
v■ = √(600/0.0035) = 414 m/s
Meeting time: t = L / (v■ + v■) = 2.00 / (490 + 414) = 2.24 × 10■³ s.
Concept: The wave with greater linear density travels slower; both pulses meet near the middle
after a very short time.
Exercise 2 – Solution
Tension: T = mg = 19.6 N
For n = 6, λ■ = 2L / n = 2(1.00)/6 = 0.333 m.
Frequency: f■ = (n/2L) √(T/µ) = (6/2) √(19.6/0.006) = 3 × 57.1 = 343 Hz.
Sound wave wavelength in air: λ■ = v■ / f■ = 343 / 343 = 1.00 m.
Concept: The 6th harmonic forms six loops (seven nodes). The same frequency of 343 Hz
generates a sound wave in air with wavelength 1.00 m.