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# CALC A guitar string is vibrating in its fundamental mode, ISBN: 9780321675460 31

## Solution for problem 74P Chapter 15

University Physics | 13th Edition

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Problem 74P

CALC A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 m. The maximum trans-verse acceleration of a point at the middle of the segment is 8.40 X 103 m/s2 and the maximum transverse velocity is 3.80 m/s. (a) What is the amplitude of this standing wave? (b) What is the wave speed for the transverse traveling waves on this string?

Step-by-Step Solution:

Solution 74P Step 1: Equation for the standing wave, y (x,t) = A sin (kx) sin (t) We know that, k x = n Or, k = n / x Provided, x = L max Therefore, k = n / L Rewriting the equation, we get, y (x,t) = A sin (nx/L) sin (t) Provided, x = L/2 (Middle of the string) and n = 1 (1st harmonics) Therefore, y (x,t) = A sin (1× [L/2]/L) sin (t) = A sin (/2) sin (t) y (x,t) = A sin (t) Velocity of the wave, v = dy/x = A cos (t) Maximum transverse velocity, v = A (Atxaximum, cos (t) = 1) Step 2: a) Velocity of the wave, v = dy/dt = A cos (t) x Differentiating again w.r.t time, we get, 2 2 2 Transverse acceleration of the wave, a = d y/dx = - A sin (t) Maximum transverse acceleration, a = - A ( x 2 At maximum, sin (t) = 1) Dividing the maximum acceleration from maximum velocity, 2 ax vx - A / A = - So, (8.40 × 10 m/s ) / 3.80 m/s = - = - 2.21 × 10 rad/s v = A x So, A = v / x A = (3.80 m/s ) / (- 2.21 × 10 rad/s) = - 0.0017 m A = - 1.7 mm Amplitude of the wave is, A = - 1.7 mm We will take the modulus of amplitude, A = 1.7 mm

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