The distance between adjacent antinodes of a standing wave on a string is 15.0 c

Question

The distance between adjacent antinodes of a standing wave on a string is 15.0 cm. A string segment at an antinode oscillates in simple harmonic motion with amplitude A = 0.85 cm and period T = 0.075 s.

i. If the string is 90.0 cm long, in which harmonic mode is the string vibrating?

ii. What are the amplitudes, wavelengths, frequencies (in Hz) and speeds of the two travelling waves that have formed the standing wave?

iii. What is the maximum transverse speed of a segment at the antinode of the standing wave?  

Explanation / Answer

(i) Adjacent nodes are also 15.0 cm apart.
(ii) Recall that cos (A B) + cos (A + B) = 2 cos A cos B, which implies that
A cos (kx – t) + A cos (kx + t) = 2 A cos (kx) cos (t). Hence the amplitude of
a standing wave is a factor two larger than the amplitude of either of the two
constituent traveling waves, but their wavelengths are the same, and their
frequencies are the same. Thus: Wavelength of each traveling wave = 30.0 cm;
Amplitude of each traveling wave A = 0.425 cm. Speed of each traveling wave v
= f = / = 30.0 cm / 0.075 s = 400 cm/s = 4 m/s.
(iii) Maximum transverse speed of a point at the antinode of the standing wave = A
= A (2 /T) = 0.850 cm (2 / 0.0750 s) = 71.21 cm/s. Minimum speed = 0.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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