A simple harmonic oscillator undergoes 19 cycles in a time of 32 s. Calculate it

Question

A simple harmonic oscillator undergoes 19 cycles in a time of 32 s. Calculate its period. 1.7 s 16.5 s 2.2 s 0.60 s 0.14 s Fifty grams of lead shot is placed in a light cylinder that is 1.6 m long and is closed at both ends. When the cylinder is placed in a vertical position and Inverted, the shot falls the length of the cylinder. Calculate the increase in temperature of the lead when this process is repented 75 times Assume no beat loss to the surroundings. (The specific heat of lead is 0.03 kcal/(kg degree C). One kcal equals 4187 J.) 22 degree C 45 degree C 15 degree C 32 degree C 9.4 degree C

Explanation / Answer

20 .) Time period = time taken to complete one cycle = 32 /19 = 1.68 ~ 1.7 sec

21.) Potential energy lost for one fall = mgh = 0.05 kg x 9.81 m/s2 x 1.6 m = 0.7848 Joules

Energy lost to heat in 75 falls = 75 x 0.7848 = 58.86 Joules

Since the cylinder is light (and mass is not given, hence neglected), we shall assume that no heat is transferred to the cylinder. Also, since no heat is lost to the surroundings, the total energy is used to raise the temperature of the lead shot.

specific heat of lead c = 0.03 kcal / Kg.C = 0.03 x 4187 Joules/ KgC = 125.61 Joules/ Kg.C

mc(T) = Energy transferred

0.05 x 125.61 x (T) = 58.86

T = 58.86 / 6.2805 = 9.37 C0 ~9.4 C0

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