Please help to solve above question. A student placed a closed loop of metal str

Question

Please help to solve above question.

A student placed a closed loop of metal string on two small roller supports at A and B which are 1 m apart and then hanged a mass of 2 kg at C directly below the centre of the two supports as shown in Figure 3. If the string has a mass of 0.1 kg/m with the angle ACB equal to 90 degrees, detemune the fundamental frequency of vibration (in Hz) for the segment of the stnng AB if it is to be excited into transverse vibration. You can assume the gravitational acceleration g to be equal to 9 81 m/s2. If the student replaced the original metal string by another closed loop of metal string of longer length and repeated the same experiment, would the fundamental frequency be higher or lower than the ongmal value? Justify your answer. Please help to solve above question.

Explanation / Answer

let the tension in the string be T.

now the horizontal component of tension in left string will balance horizontal component of tension in right string.

now to balance the vertical force:

(as angle ACB isn 90 degree and AC=AB, then angle with vertical made by either string=45 degree)

2*T*cos(45)=2*9.81

T=13.873 N

so speed of the wave=sqrt(tension/mass per unit length)=11.78 m/s

assuming fundamnetal frequency to be calculated, wavelength=0.5*length of the string=0.5 m

as we know wavelength*frequency=speed of the wave

frequency=23.56 Hz

b)as speed depends upon mass per unit length, changing the length wont affect it

but as wavelength depends upon the length of the string,

if you increase the length, wavelength will increase.

then freqeuency=speed/wavelength will decrease.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.