Question

A sinusoidal wave travels along a stretched string. A particle on the string has a maximum speed of 2.0 m/s and a maximum acceleration of 200 m/s2. What is the amplitude of the wave?

Answer 1

The amplitude of the sinusoidal wave can be calculated by rearranging the simple harmonic motion formulas for maximum speed and acceleration, resulting in an amplitude of 0.02 meters.

Step-by-step explanation:

To calculate the amplitude of a sinusoidal wave where a particle on the string has a maximum speed and maximum acceleration, one can use the relationship between acceleration, velocity, and amplitude in simple harmonic motion. The maximum acceleration (a_max) is proportional to the square of the angular frequency (ω) times the amplitude (A), given by the equation a_max = ω^2 × A. Similarly, the maximum speed (v_max) is proportional to the angular frequency times the amplitude, given by v_max = ω × A.

By solving these equations simultaneously, one can eliminate the angular frequency ω to find the amplitude. The equations are A = v_max / ω and A = a_max / ω^2. When we rearrange these equations and solve for ω in terms of v_max and substitute back into the acceleration equation, we get A = v_max^2 / a_max. With a maximum speed of 2.0 m/s and a maximum acceleration of 200 m/s2, the amplitude can be calculated as follows: A = (2.0 m/s)^2 / (200 m/s2) = 0.02 m, which is the amplitude of the wave.

Answer 2

The amplitude of the wave is 0.02 m.

Step-by-step explanation:

Given that,

Maximum speed = 2.0 m/s

Maximum acceleration = 200 m/s²

We need to calculate the angular frequency

Using formula of angular frequency

`\omega=(a_(max))/(v_(max))`

Put the value into the formula

`\omega=(200)/(2.0)`

`\omega=100\ rad/s`

We need to calculate the amplitude of the wave

Using formula of velocity

`v_(max)=A\omega`

`A=(v_(max))/(\omega)`

Put the value into the formula

`A=(2.0)/(100)`

`A=0.02\ m`

Hence, The amplitude of the wave is 0.02 m.