(a) What is the frequency of the vibrator?(Note: The greater the tension in the

Question

(a) What is the frequency of the vibrator?(Note: The greater the tension in the string, the smaller thenumber of nodes in the standing wave.)
Hz

(b) What is the largest object mass for which standing waves couldbe observed?
kg Hz

(b) What is the largest object mass for which standing waves couldbe observed?
kg 9) In the arrangement shown below, an object can be hung from a string (with a linear mass density mu = 0.00200 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L =2.30 m. When the mass m of the object is either 16.0 kg or 25.0 kg, standing waves are observed; however, no standing waves are observed with any mass between these values. (a) What is the frequency of the vibrator?(Note: The greater the tension in the string, the smaller the number of nodes in the standing wave.) Hz (b) What is the largest object mass for which standing waves couldbe observed? kg

Explanation / Answer

we know that     f = n v /2L     for standing waves and .         v = tension/ = mg /   = m * 9.80 / 0.002    =   70m . So now... .          f = n * 70m /2*2.30        .       f = n * 15.2174 m       . for the two values of m, we get .         f = n1 * 15.2174 *16                f = n1 * 60.87 .        f = n2 *15.2174 *25                 f = n2 * 76.087 . These must be adjacent resonances, so n1 = n2+ 1        and .             f = (n2 + 1) *60.87            f = n2 * 76.087   .          f  = (f/76.087 + 1) *60.87            76.087 f   = f * 60.87   + 76.087 *60.87 .           15.2174f    =    4631.38 .       f =   304.35   Hz    is thefreq of the oscillator . for this freq, the fastest speed that can form a standing waveis .          v =2Lf/n = 2 * 2.30 * 304.35 / 1 =   1400 m/s . And    v2 = tension /       so          tension = v2 = 0.002 *14002   =    3920 Newtons . So     mass = weight / g = 3920 / 9.80   =    400 kg

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