A transverse wave on a string is described by the equation y(x,t) = 2.20 cm sin(130 rad s⁻¹ t + 15 rad m⁻¹ x).

• I. Find the approximate maximum transverse speed of a point on the string.
• (A) 1.2 m s⁻¹
• (B) 1.7 m s⁻¹
• (C) 2.9 m s⁻¹
• (D) 3.4 m s⁻¹
• II. Find the approximate maximum transverse acceleration of a point on the string.
• (A) 300 m s⁻²
• (B) 372 m s⁻²
• (C) 410 m s⁻²
• (D) 450 m s⁻²
• III. Find the approximate speed of a wave moving along the string.
• (A) 4.2 m s⁻¹
• (B) 5.6 m s⁻¹
• (C) 7.4 m s⁻¹
• (D) 8.7 m s⁻¹

The wave on the string is described by the equation y(x,t) = 2.20 cm sin(130 rad s⁻¹ t + 15 rad m⁻¹ x). Let's break down the calculations step by step.

Maximum Transverse Speed

The maximum transverse speed (v_max) can be found using the formula:

v_max = Aω

Where:

• A = amplitude = 2.20 cm = 0.022 m
• ω = angular frequency = 130 rad/s

Calculating:

v_max = 0.022 m * 130 rad/s = 2.86 m/s

The closest option is (C) 2.9 m/s.

Maximum Transverse Acceleration

The maximum transverse acceleration (a_max) is calculated using:

a_max = Aω²

Substituting the values:

a_max = 0.022 m * (130 rad/s)² = 0.022 m * 16900 = 372.8 m/s²

The nearest choice is (B) 372 m/s².

Wave Speed

The speed of the wave (v) can be determined with the formula:

v = fλ

First, find the wave frequency (f) using:

f = ω / (2π) = 130 rad/s / (2π) ≈ 20.7 Hz

Next, calculate the wavelength (λ):

λ = 2π/k = 2π / 15 rad/m ≈ 0.42 m

Now, calculate the wave speed:

v = fλ ≈ 20.7 Hz * 0.42 m ≈ 8.69 m/s

The closest answer is (D) 8.7 m/s.

In summary:

• Maximum Transverse Speed: (C) 2.9 m/s
• Maximum Transverse Acceleration: (B) 372 m/s²
• Wave Speed: (D) 8.7 m/s