Question

2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If the speed of waves on the string is 150 m/s, what is the maximum speed for a point on the string at an antinode?

Answer 1

Answer:

Maximum speed for a point on the string at anti node will be 22.6 m/sec

Explanation:

We have given length of string L = 3.5 m

For 7th harmonic length of the string $L=\frac{7\lambda }{2}$

So $\lambda =\frac{2L}{7}$

Speed of the wave in the string is 150 m/sec

Frequency corresponding to this wavelength $f=\frac{v}{\lambda }=\frac{7v}{2L}$

So angular frequency will be equal to $\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec$

Maximum speed is equal to $v_m=A\omega =0.024\times 942=22.60m/sec$

So maximum speed for a point on the string at anti node will be 22.6 m/sec