In Fig. 16-41, a string, tied to a sinusoidal oscillator at P and running over a

Question

In Fig. 16-41, a string, tied to a sinusoidal oscillator at P and running over a support at Q, is stretched by a block of mass m. The separation L between P and Q is 1.20 m, and the frequency f of the oscillator is fixed at 120 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging block is 286.1 g or 447.0 g, but not for any intermediate mass. What is the linear density of the string?

Explanation / Answer

L = 1.20 m => 2 x L = 2.4 m f = 120 Hz m = 286.1 g & 447.0 g Let u is mass/lenght T1 = 0.2861 x 9.8 = 2.80378 N T2 = 0.447 x 9.8 = 4.3806 N f = n/(2L) sqrt(T/u) 120 = n/2.4 sqrt(2.80378/u) 120=(n-1)/2.4 sqrt(4.3806/u) On solving we get, n x 1.67 = (n-1) x 2.09 => n = 4.975 , Thus n = 5 ( on approximation). Therefore , putting back this value in any of the above equations, we get, 120 = 5/2.4 x sqrt(2.80378/u) => u = 8.45 x 10^-4 kg/m = 0.845 g/m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---