Elast 6.1 n he arrangement shewn below, an obyect can be hung fan a string (with

Question

Elast 6.1 n he arrangement shewn below, an obyect can be hung fan a string (with linear mas density u as 0.0015 kg/m) that p The string is or constant frequency f, and the length of the string between pent P and the pulley is 2.4 m. when the mass m of the object is eilher2(3Dkg the30.0 kgox 2KX kg, s observed; no standing waves are observed with any mass between these s values what is he frequency of the vibrator? [Note that this pecture is meant to be illustrative, tx necessarilyshow the relevant number of nodesants-nodes for this literal it do es problem.] Vibrant a, 451.8 Hz b. 301.2Hz c. 753.1 Hz d. 361.5Hz c. 96.2 Hz dent n a senes his Determi

Explanation / Answer

µ = 0.0015 kg/m,  L = 2.4 m. m1 = 20 kg, m2 =28.8 kg,

v = (mg/)

wavelength : L = n(/2), = 2L/n

frequency f = v/ = n(mg/)/(2L) =constant

n1m1 =n2m2 = constant

5n1 = 6n2

we have n1 = 6, n2 = 5

f = 6(m1g/)/(2L) = 989.95 Hz

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