# For a standing wave on a string, the distance between nodes is 0.125 mm, the frequency is 296 Hz,...

## Question:

For a standing wave on a string, the distance between nodes is 0.125 mm, the frequency is 296 Hz, and the amplitude is {eq}1.40 \times10^{-3}\ m {/eq}.

a) What is the speed of waves on this string.

b) What is the maximum transverse velocity at an antinode?

c) What is the maximum transverse acceleration at an antinode?

## Maximum Acceleration in Simple Harmonic Motion:

We know that the acceleration of a particle executing a simple harmonic motion is directly proportional to the position of the particle relative to its equilibrium position and is always directed towards the equilibrium position. Then, the maximum acceleration of the particle in simple harmonic motion will occur when it is at the maximum distance from the mean position.

The magnitude of the maximum acceleration {eq}(a_{max}) {/eq} of a particle executing a simple harmonic motion is equal to the square of the angular frequency (ω) of the motion multiplied with the amplitude (A) of the simple harmonic motion and can be expressed as:

{eq}\displaystyle{\hspace{8cm}a_{max} = \omega^2A} {/eq}

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We are given

• The distance between the two successive nodes: {eq}d = 0.125 \ \rm mm {/eq}
• The frequency of the wave: {eq}f = 296 \ \rm Hz {/eq}
• Th...