A string with both ends held fixed is vibrating in its third harmonic. The waves

Question

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 235 Hz . The amplitude of the standing wave at an antinode is 0.380 cm .

A.)

Calculate the amplitude at point on the string a distance of 18.0 cm from the left-hand 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 at this point?

C.)

Calculate the maximum transverse velocity of the string at this point.

D.)

Calculate the maximum transverse acceleration of the string at this point.

Explanation / Answer

we know the wavelength is given by the equation

=c/

therefore =193/235

=0.821 metres

equation of standing wave is

y(x,t)=2Asin(kx) cos(t)

where =2** and k=2*/

amplitute at antinode is=0.38 cm=0.0038 m

this is the maximum amplitute which is nothing but 2A

therefore 2 *A=0.0038

also k=2 * /

k=(2 * 3.14)/0.821=7.6492

A)amplitute at the distance 18 cm from the left hand

=2A sin(k * x)

=0.0038 * sin(7.6492 *0.18)

=0.0038 *0.98125 m

=0.003728 m

=0.3728 cms

B)Time to go from largest upward to largest downwards

is T/2

=1/(2 *)

=1/(2 * 235)

=0.002127 secs

C) maximum transverse velocity of the string at this point.

V at pt =2A * * sin( k * x) [ w= 2 * * ]

v=0.0038 * 0.98125 * 2 * * 235

v=5.5028 m/s

D) maximum transverse acceleration of the string at this point

a=2A * * *  sin( k * x)

a=5.50 * 2 * 3.14 * 235

a=8116.9 m/s2

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