Find the oscillation amplitude

[NivekOh]

Homework Statement

A 2.0 m long string vibrates at its second harmonic frequency with a maximum amplitude of 2.0cm. One end of the string is at x = 0cm. Find the oscillation amplitude at x = 10,20,30,40, and 50cm.

Homework Equations

A(x) = 2asin(kx)
k= 2pi/wavelength

The Attempt at a Solution

I tried plugging the numbers into the equatio, but couldn't get the right answer. Can someone get me on the right track?

 

[flatmaster]

I'm not sure why you have a two in your equation for Amplitude. Because you're on the second harmonic, the total 2m leingth of the string is your wavelength. The 2.0 cm is your amplitued a. Plug and solve without the 2.
Also, make sure you use consistant units. Change your cm to meters.

 

[thomate1]

The oscillation amplitude at a specific point on a string is given by the equation A(x) = 2asin(kx), where A(x) is the amplitude at position x, a is the maximum amplitude, and k is the wave number, given by k = 2π/λ, where λ is the wavelength.

In this problem, we are given that the string has a length of 2.0 m and is vibrating at its second harmonic frequency, which means that the wavelength is equal to half the length of the string, or 1.0 m.

To find the wave number, we can substitute the wavelength into the equation for k, giving us k = 2π/1.0 = 2π.

Now, we can use this value for k to find the oscillation amplitude at different points on the string.

At x = 10 cm, the oscillation amplitude would be A(10) = 2asin(2π(10)/100) = 0.4 cm.

Similarly, at x = 20 cm, the amplitude would be A(20) = 2asin(2π(20)/100) = 0.8 cm.

At x = 30 cm, the amplitude would be A(30) = 2asin(2π(30)/100) = 1.2 cm.

At x = 40 cm, the amplitude would be A(40) = 2asin(2π(40)/100) = 1.6 cm.

And finally, at x = 50 cm, the amplitude would be A(50) = 2asin(2π(50)/100) = 2.0 cm, which is the maximum amplitude given in the problem.

So, the oscillation amplitude increases as we move away from the fixed end of the string at x = 0 cm, reaching its maximum value at the midpoint of the string and then decreasing again as we move towards the other end.