Question
Part A. Here on Earth, a simple pendulum of length L isobserved to have a frequency f. If another pendulum on Earthis observed to have a frequency that is 23f, what is itslenght? A. L/4.796 B.23L C.529L D. L/23 E. 4.796L F. L/529 G. L/1058 H. 1058L Part B. Here on Earth, a simple pendulum of length L isobserved to have a frequency f. On another planet a pendulum oflength L is observed to have a frequency of 97f. What is theacceleration due to gravity on this planet? In the answers below,g= 9.8m/s2. A. 97g B. g/ 18818 C. 9.849g D.g/97 E.g/9.849 F. g/9409 G. 9409g H.18818g Part A. Here on Earth, a simple pendulum of length L isobserved to have a frequency f. If another pendulum on Earthis observed to have a frequency that is 23f, what is itslenght? A. L/4.796 B.23L C.529L D. L/23 E. 4.796L F. L/529 G. L/1058 H. 1058L Part B. Here on Earth, a simple pendulum of length L isobserved to have a frequency f. On another planet a pendulum oflength L is observed to have a frequency of 97f. What is theacceleration due to gravity on this planet? In the answers below,g= 9.8m/s2. A. 97g B. g/ 18818 C. 9.849g D.g/97 E.g/9.849 F. g/9409 G. 9409g H.18818g
Explanation / Answer
. (a) . Frequency of the first pendulum , f1 = f . Length of the first pendulum, L1 = L . Frequencyof second pendulum, f2 = 23 f . Lengthof the second pendulum, L2 = ? . Frequency = ( 1/ 2 ) ( g / L ) . f 1 / L . L ( 1 / f 2) . L2 / L1 = ( f2 / f1 ) 2 . L2 = 23 2 * L1 . = 529 L . Ans:Option - C . (b) . Frequency of pendulum on earth, f1 = f . Frequency of pendulum on other planet, f2 = 97 f . Acceleration due to gravity on earth, g1 = g . Acceleration due to gravity on the planet, g2 = ? . Frequency, f g . g f 2 ( when L = constant) . (g2 / g1 ) = ( f2 / f1 ) 2 . g2 = 97 2 * g1 . g2 = 9409 g . Ans: option - G . g2 = 97 2 * g1 . g2 = 9409 g . Ans: option - G .