. A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m>s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement

Explanation:

The formula holding the relationship between wavelength and frequency of a wave can be expressed as:

where;

wavelength

v = speed

f = frequency

Given that:

v = 192 m/s and f = 240Hz

Then;

Now, to estimate the respective amplitude of the string, we need to approach it by using the concept of wave equation which is:

y = A sin kx

here;

A = amplitude of the standing wave

k = wave number

x = maximum displacement

y = distance from center

recall that:

∴

Now;

for A = 0.400 cm ; = 0.800 m ; k = 40 cm

Then;

y = 0.400 sin π

y = 0 cm

At distance 40 cm; the amplitude = 0 cm

Thus, it is a node.

For k = 20cm

Then:

y = 0.400 sin π/2

y = 0.400 cm

At distance 20 cm; the amplitude = 0.400 cm

Thus, it is antinode.

For k = 10cm

Then:

4y = 0.400 sin π/2

y = 0.283 cm

At distance 10 cm; the amplitude = 0.283 cm

b)

The required time taken to go through the displacement( i.e. from largest upward to downward) is the time required to cover half of the wavelength.

This is expressed as:

T = 0.00208

T = 2.08 × 10⁻³ s

the answer is b

explanation: