A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 195 m/s and a frequency of 220 Hz. The amplitude of the standing wave at an antinode is 0.430 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) How much time does it take the string to go from its largest upward displacement to its largest downward displacement? C) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at this point.

a) i) 0.18cm ii)  0.09cm iii) 0.045cm b) 0.96s c) v= 594 m/s a= 8.2*10⁵ m²/s

Explanation:

a) We have to find the amplitude at distance of i) 40 cm ii) 20 cm iii) 10 cm.

Now, the amplitude is related to frequency by relation

Distance= Amplitude * Frequency

The frequency given = 220 Hz

So for

i) 40= Amplitude * 220 Hz

Amplitude= = 0.18 cm

ii) 20= Amplitude * 220 Hz

Amplitude= = 0.09 cm

iii) 10= Amplitude * 220 Hz

Amplitude= = 0.045 cm

b) Wavelength= Velocity/ Frequency

Wavelength= 195/220 = 0.89m

To find the time from its upward largest displacement to downard largest displacement. We multiply the amplitude by 2 and divide by wavelength

T= 2*0.43/0.89

T= 0.96 s

c) To find the maximum transverse velocity it is equal to angular velocity times amplitude

So Velocity= = = 594 m/s

Acceleration is transverse velocity times angular velocity= = = 8.2*10⁵ m²/s

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6