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m=(kT^(2))/(2pi) An experimental physicist uses the given equation to model the relationship between an m kilogram mass connected to a spring with spring constant, k, and a period of oscillation, T. If the physicist changes the mass and the period of oscillation drops to (1)/(4) of its original value, which of the following describes the change in mass? Choose 1 answer: (A) The new mass is (1)/(2) of the original mass. (B) The new mass is (1)/(4) of the original mass. (C) The new mass is (1)/(16) of the original mass. (D) The new mass is quadrupled.

m=kT22π m=\frac{k T^{2}}{2 \pi} \newlineAn experimental physicist uses the given equation to model the relationship between an m m kilogram mass connected to a spring with spring constant, k k , and a period of oscillation, T T . If the physicist changes the mass and the period of oscillation drops to 14 \frac{1}{4} of its original value, which of the following describes the change in mass?\newlineChoose 11 answer:\newline(A) The new mass is 12 \frac{1}{2} of the original mass.\newline(B) The new mass is 14 \frac{1}{4} of the original mass.\newline(C) The new mass is 116 \frac{1}{16} of the original mass.\newline(D) The new mass is quadrupled.

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Q. m=kT22π m=\frac{k T^{2}}{2 \pi} \newlineAn experimental physicist uses the given equation to model the relationship between an m m kilogram mass connected to a spring with spring constant, k k , and a period of oscillation, T T . If the physicist changes the mass and the period of oscillation drops to 14 \frac{1}{4} of its original value, which of the following describes the change in mass?\newlineChoose 11 answer:\newline(A) The new mass is 12 \frac{1}{2} of the original mass.\newline(B) The new mass is 14 \frac{1}{4} of the original mass.\newline(C) The new mass is 116 \frac{1}{16} of the original mass.\newline(D) The new mass is quadrupled.
  1. Denote Original Mass and Period: Let's denote the original mass as mm and the original period of oscillation as TT. According to the given equation, we have:\newlinem=kT22πm = \frac{kT^2}{2\pi}
  2. Calculate New Period: Now, let's denote the new period of oscillation as T′T'. Given that T′T' is one-fourth of the original period TT, we can write:\newlineT′=T/4T' = T / 4
  3. Find New Mass with New Period: We need to find the new mass m′m' using the same equation with the new period T′T'. So we have:\newlinem′=k(T′)22πm' = \frac{k(T')^2}{2\pi}
  4. Substitute New Period: Substitute T′=T4T' = \frac{T}{4} into the equation for m′m':\newlinem′=k(T4)22πm' = \frac{k\left(\frac{T}{4}\right)^2}{2\pi}
  5. Simplify Equation: Simplify the equation by squaring T4\frac{T}{4}:m′=k(T216)2πm' = \frac{k\left(\frac{T^2}{16}\right)}{2\pi}
  6. Compare New Mass with Original Mass: Now, we can compare the new mass m′m' with the original mass mm. We can see that m′m' is 116\frac{1}{16} of mm because (T2/16)(T^2/16) is 116\frac{1}{16} of T2T^2: m′=m16m' = \frac{m}{16}
  7. Conclusion: Therefore, the new mass is 116\frac{1}{16} of the original mass. This corresponds to answer choice (C)(C).

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