201. At what displacement magnitude is the kinetic energy of a spring oscillator equal to its potential energy?
ⓐ. \(\frac{A}{2}\)
ⓑ. \(\frac{A}{\sqrt{2}}\)
ⓒ. \(\frac{\sqrt{3}A}{2}\)
ⓓ. \(A\)
Correct Answer: \(\frac{A}{\sqrt{2}}\)
Explanation: \( \textbf{Energy condition:} \) Kinetic energy equals potential energy, so \(K=U\).
The total energy is
\[
E=K+U
\]
If \(K=U\), then each is half the total energy:
\[
U=\frac{E}{2}
\]
For spring SHM,
\[
U=\frac{1}{2}kx^2
\]
and
\[
E=\frac{1}{2}kA^2
\]
So,
\[
\frac{1}{2}kx^2=\frac{1}{2}\left(\frac{1}{2}kA^2\right)
\]
Cancel \(\frac{1}{2}k\):
\[
x^2=\frac{A^2}{2}
\]
Taking magnitude:
\[
|x|=\frac{A}{\sqrt{2}}
\]
At this displacement, half the energy is kinetic and half is stored as spring potential energy.
\( \textbf{Final answer:} \) \(|x|=\frac{A}{\sqrt{2}}\).
202. A spring oscillator has total energy \(0.80\,\text{J}\). At one instant its potential energy is \(0.30\,\text{J}\). The kinetic energy at that instant is
ⓐ. \(0.24\,\text{J}\)
ⓑ. \(0.30\,\text{J}\)
ⓒ. \(0.50\,\text{J}\)
ⓓ. \(1.10\,\text{J}\)
Correct Answer: \(0.50\,\text{J}\)
Explanation: \( \textbf{Given:} \) Total energy \(E=0.80\,\text{J}\), potential energy \(U=0.30\,\text{J}\).
In ideal SHM, total mechanical energy is the sum of kinetic and potential energies:
\[
E=K+U
\]
Rearrange for kinetic energy:
\[
K=E-U
\]
Substitute the values:
\[
K=0.80\,\text{J}-0.30\,\text{J}
\]
\[
K=0.50\,\text{J}
\]
This result assumes ideal SHM, where no mechanical energy is lost during the motion.
The kinetic and potential energies change separately, but their sum remains constant.
\( \textbf{Final answer:} \) \(K=0.50\,\text{J}\).
203. A spring oscillator of mass \(0.50\,\text{kg}\) has speed \(1.2\,\text{m s}^{-1}\) at a certain instant. Its kinetic energy then is
ⓐ. \(0.18\,\text{J}\)
ⓑ. \(0.30\,\text{J}\)
ⓒ. \(0.36\,\text{J}\)
ⓓ. \(0.72\,\text{J}\)
Correct Answer: \(0.36\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(m=0.50\,\text{kg}\), \(v=1.2\,\text{m s}^{-1}\).
Kinetic energy is
\[
K=\frac{1}{2}mv^2
\]
Substitute the given values:
\[
K=\frac{1}{2}(0.50)(1.2)^2
\]
First square the speed:
\[
(1.2)^2=1.44
\]
Now calculate:
\[
K=0.25(1.44)
\]
\[
K=0.36\,\text{J}
\]
The square of speed is required, so halving or doubling speed would not change kinetic energy linearly.
\( \textbf{Final answer:} \) \(K=0.36\,\text{J}\).
204. The total energy of an ideal spring oscillator is
ⓐ. \(E=\frac{1}{2}kA^2\)
ⓑ. \(E=\frac{1}{2}kx^2\) at every point only
ⓒ. \(E=kA\)
ⓓ. \(E=\frac{1}{2}mA^2\)
Correct Answer: \(E=\frac{1}{2}kA^2\)
Explanation: The total energy of an ideal spring oscillator is equal to its maximum potential energy at an extreme position. At the extreme, \(|x|=A\), and the speed is zero. The spring potential energy there is \(U_{\max}=\frac{1}{2}kA^2\). Since total energy remains constant in ideal SHM, \(E=\frac{1}{2}kA^2\) at every position. The expression \(\frac{1}{2}kx^2\) gives potential energy at a general displacement, not the total energy unless \(|x|=A\).
205. A spring oscillator has \(m=0.20\,\text{kg}\), \(\omega=10\,\text{rad s}^{-1}\), and amplitude \(A=0.050\,\text{m}\). Its total energy is
ⓐ. \(0.0125\,\text{J}\)
ⓑ. \(0.025\,\text{J}\)
ⓒ. \(0.050\,\text{J}\)
ⓓ. \(0.100\,\text{J}\)
Correct Answer: \(0.025\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(m=0.20\,\text{kg}\), \(\omega=10\,\text{rad s}^{-1}\), \(A=0.050\,\text{m}\).
For SHM, total energy may be written as
\[
E=\frac{1}{2}m\omega^2A^2
\]
This form is useful when \(m\), \(\omega\), and \(A\) are known.
Substitute the values:
\[
E=\frac{1}{2}(0.20)(10)^2(0.050)^2
\]
Calculate each part:
\[
\frac{1}{2}(0.20)=0.10
\]
\[
(10)^2=100
\]
\[
(0.050)^2=0.0025
\]
Now multiply:
\[
E=0.10\times100\times0.0025
\]
\[
E=0.025\,\text{J}
\]
The amplitude is squared, so using \(0.050\) without squaring would overestimate the energy.
\( \textbf{Final answer:} \) \(E=0.025\,\text{J}\).
206. If the amplitude of an ideal spring oscillator is doubled while \(k\) remains unchanged, the total energy becomes
ⓐ. twice the original energy value
ⓑ. four times the original energy
ⓒ. half the original energy value
ⓓ. same as the original energy value
Correct Answer: four times the original energy
Explanation: The total energy of an ideal spring oscillator is \(E=\frac{1}{2}kA^2\). If \(k\) is unchanged, energy is proportional to \(A^2\). Doubling amplitude changes \(A\) to \(2A\). The new energy is \(E'=\frac{1}{2}k(2A)^2=4\left(\frac{1}{2}kA^2\right)\). Thus, the energy becomes four times as large. The square dependence means amplitude changes have a stronger effect on energy than a simple linear comparison suggests.
207. Two identical spring oscillators have amplitudes \(A\) and \(3A\). The ratio of their total energies is
ⓐ. \(1:3\)
ⓑ. \(1:6\)
ⓒ. \(1:9\)
ⓓ. \(3:1\)
Correct Answer: \(1:9\)
Explanation: \( \textbf{Given:} \) The spring oscillators are identical, so they have the same \(k\).
Total energy is
\[
E=\frac{1}{2}kA^2
\]
For the first oscillator:
\[
E_1=\frac{1}{2}kA^2
\]
For the second oscillator with amplitude \(3A\):
\[
E_2=\frac{1}{2}k(3A)^2
\]
\[
E_2=\frac{1}{2}k(9A^2)
\]
\[
E_2=9E_1
\]
Therefore,
\[
E_1:E_2=1:9
\]
The amplitude ratio is squared when comparing energies.
\( \textbf{Final answer:} \) \(E_1:E_2=1:9\).
208. Use the graph description below.
For an ideal spring oscillator, \(U-x\), \(K-x\), and total energy \(E-x\) graphs are drawn on the same axes.
The total energy graph should appear as
ⓐ. a horizontal line
ⓑ. a parabola opening upward through the origin
ⓒ. a parabola opening downward touching zero at \(x=0\)
ⓓ. a straight line with negative slope
Correct Answer: a horizontal line
Explanation: In ideal SHM, the total mechanical energy does not depend on the instantaneous displacement \(x\). For a spring oscillator, \(U=\frac{1}{2}kx^2\) and \(K=\frac{1}{2}k(A^2-x^2)\). Their sum is
\[
E=U+K=\frac{1}{2}kA^2
\]
This is constant for all allowed positions from \(-A\) to \(+A\). Therefore, the \(E-x\) graph is a horizontal line. The separate \(U-x\) and \(K-x\) graphs curve in opposite ways, but their sum stays fixed.
209. The kinetic energy graph \(K\) versus displacement \(x\) for ideal spring SHM has the shape of
ⓐ. an upward-opening parabola with minimum at \(x=0\)
ⓑ. a downward-opening parabola with maximum at \(x=0\)
ⓒ. a straight line through the origin with positive slope
ⓓ. a horizontal line showing constant kinetic energy
Correct Answer: a downward-opening parabola with maximum at \(x=0\)
Explanation: Kinetic energy in spring SHM is \(K=\frac{1}{2}k(A^2-x^2)\). This is a quadratic expression in \(x\) with a negative coefficient of \(x^2\). Therefore, the \(K-x\) graph is a downward-opening parabola. It is maximum at \(x=0\), where \(K=\frac{1}{2}kA^2\). It becomes zero at \(x=+A\) and \(x=-A\). This graph is complementary to the \(U-x\) graph, which opens upward.
210. A spring oscillator has \(E=2.0\,\text{J}\). At a certain displacement, its kinetic energy is three times its potential energy. The potential energy at that instant is
ⓐ. \(0.25\,\text{J}\)
ⓑ. \(0.50\,\text{J}\)
ⓒ. \(1.00\,\text{J}\)
ⓓ. \(1.50\,\text{J}\)
Correct Answer: \(0.50\,\text{J}\)
Explanation: \( \textbf{Given:} \) Total energy \(E=2.0\,\text{J}\), and \(K=3U\).
Total energy is
\[
E=K+U
\]
Substitute \(K=3U\):
\[
E=3U+U
\]
\[
E=4U
\]
So,
\[
U=\frac{E}{4}
\]
Substitute \(E=2.0\,\text{J}\):
\[
U=\frac{2.0}{4}\,\text{J}
\]
\[
U=0.50\,\text{J}
\]
Then \(K=1.50\,\text{J}\), and the two add back to \(2.0\,\text{J}\).
\( \textbf{Final answer:} \) \(U=0.50\,\text{J}\).
211. A spring oscillator has total energy \(E\). When \(x=\frac{A}{2}\), the kinetic energy is
ⓐ. \(\frac{E}{4}\)
ⓑ. \(\frac{2E}{4}\)
ⓒ. \(\frac{3E}{4}\)
ⓓ. \(\frac{4E}{4}\)
Correct Answer: \(\frac{3E}{4}\)
Explanation: \( \textbf{Total energy:} \)
\[
E=\frac{1}{2}kA^2
\]
Potential energy at displacement \(x\) is
\[
U=\frac{1}{2}kx^2
\]
At \(x=\frac{A}{2}\),
\[
U=\frac{1}{2}k\left(\frac{A}{2}\right)^2
\]
\[
U=\frac{1}{2}k\frac{A^2}{4}
\]
\[
U=\frac{1}{4}\left(\frac{1}{2}kA^2\right)
\]
\[
U=\frac{E}{4}
\]
Since \(K=E-U\),
\[
K=E-\frac{E}{4}=\frac{3E}{4}
\]
At half-amplitude displacement, most of the energy is still kinetic.
\( \textbf{Final answer:} \) \(K=\frac{3E}{4}\).
212. A compact energy record for a spring oscillator says:
I. \(U\) is maximum at \(|x|=A\).
II. \(K\) is maximum at \(x=0\).
III. \(E\) changes with \(x\) in ideal SHM.
Select the valid set.
ⓐ. I only
ⓑ. I, II, and III
ⓒ. II and III only
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Spring potential energy is \(U=\frac{1}{2}kx^2\), so it is maximum at the extreme positions where \(|x|=A\). Kinetic energy is maximum at the mean position because the speed is maximum there. In ideal SHM, total mechanical energy remains constant and does not change with \(x\). Therefore, statement III is not valid. The separate energy forms vary with position, but their sum stays fixed.
213. A spring oscillator has \(k=50\,\text{N m}^{-1}\) and total energy \(1.0\,\text{J}\). Its amplitude is
ⓐ. \(0.10\,\text{m}\)
ⓑ. \(0.40\,\text{m}\)
ⓒ. \(0.30\,\text{m}\)
ⓓ. \(0.20\,\text{m}\)
Correct Answer: \(0.20\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(k=50\,\text{N m}^{-1}\), \(E=1.0\,\text{J}\).
For an ideal spring oscillator,
\[
E=\frac{1}{2}kA^2
\]
Rearrange for \(A^2\):
\[
A^2=\frac{2E}{k}
\]
Substitute:
\[
A^2=\frac{2(1.0)}{50}
\]
\[
A^2=0.040
\]
Now take the positive square root:
\[
A=\sqrt{0.040}
\]
\[
A=0.20\,\text{m}
\]
Amplitude is reported as a positive magnitude, even though displacement can be positive or negative during the motion.
\( \textbf{Final answer:} \) \(A=0.20\,\text{m}\).
214. A spring oscillator and another spring oscillator have the same amplitude, but the second spring has twice the force constant of the first. The ratio of their total energies \(E_1:E_2\) is
ⓐ. \(1:1\)
ⓑ. \(1:4\)
ⓒ. \(2:1\)
ⓓ. \(1:2\)
Correct Answer: \(1:2\)
Explanation: \( \textbf{Energy relation:} \)
\[
E=\frac{1}{2}kA^2
\]
Both oscillators have the same amplitude \(A\).
Let the first spring constant be \(k\), so
\[
E_1=\frac{1}{2}kA^2
\]
The second spring has force constant \(2k\), so
\[
E_2=\frac{1}{2}(2k)A^2
\]
\[
E_2=kA^2
\]
Compare:
\[
E_2=2E_1
\]
Therefore,
\[
E_1:E_2=1:2
\]
For the same amplitude, a stiffer spring stores more energy because its force rises more strongly with displacement.
\( \textbf{Final answer:} \) \(E_1:E_2=1:2\).
215. On the same displacement axis for an ideal spring oscillator, the \(U-x\), \(K-x\), and \(E-x\) graphs are compared. The best description is
ⓐ. \(U\): upward parabola; \(K\): downward parabola; \(E\): horizontal line
ⓑ. \(U\): downward parabola; \(K\): upward parabola; \(E\): horizontal line
ⓒ. \(U\): horizontal line; \(K\): upward parabola; \(E\): downward parabola
ⓓ. \(U\): upward parabola; \(K\): horizontal line; \(E\): downward parabola
Correct Answer: \(U\): upward parabola; \(K\): downward parabola; \(E\): horizontal line
Explanation: In an ideal spring oscillator, spring potential energy is \(U=\frac{1}{2}kx^2\), so its graph against \(x\) is an upward-opening parabola. Kinetic energy is \(K=\frac{1}{2}k(A^2-x^2)\), so its graph against \(x\) is a downward-opening parabola. The total energy is \(E=\frac{1}{2}kA^2\), which is independent of \(x\), so its graph is a horizontal line. The \(U\) and \(K\) graphs exchange height as \(x\) changes, but their sum remains fixed. This graph set is a visual form of conservation of mechanical energy in ideal SHM.
216. Use the graph description below.
For a spring oscillator, curve P is zero at \(x=0\) and maximum at \(x=\pm A\). Curve Q is maximum at \(x=0\) and zero at \(x=\pm A\). Curve R is a horizontal line between \(x=-A\) and \(x=+A\).
The curves P, Q, and R represent respectively
ⓐ. \(K\), \(U\), and \(E\)
ⓑ. \(U\), \(K\), and \(E\)
ⓒ. \(E\), \(K\), and \(U\)
ⓓ. \(U\), \(E\), and \(K\)
Correct Answer: \(U\), \(K\), and \(E\)
Explanation: Spring potential energy is \(U=\frac{1}{2}kx^2\), so it is zero at \(x=0\) and maximum at \(x=\pm A\). Kinetic energy is maximum at the mean position and zero at the extremes, so curve Q represents \(K\). Total energy remains constant in ideal SHM, so curve R represents \(E\). The descriptions of P and Q are complementary because energy changes from kinetic form to potential form and back. The horizontal energy line shows that the exchange occurs without mechanical energy loss in the ideal case.
217. If \(x=A\sin\omega t\) for a spring oscillator, the potential energy varies with time as
ⓐ. \(U=\frac{1}{2}kA^2\sin^2\omega t\)
ⓑ. \(U=\frac{1}{2}kA^2\sin\omega t\)
ⓒ. \(U=\frac{1}{2}kA\sin^2\omega t\)
ⓓ. \(U=\frac{1}{2}mA^2\cos^2\omega t\)
Correct Answer: \(U=\frac{1}{2}kA^2\sin^2\omega t\)
Explanation: \( \textbf{Given displacement:} \)
\[
x=A\sin\omega t
\]
For a spring oscillator, potential energy is
\[
U=\frac{1}{2}kx^2
\]
Substitute the displacement expression:
\[
U=\frac{1}{2}k(A\sin\omega t)^2
\]
Square both factors:
\[
U=\frac{1}{2}kA^2\sin^2\omega t
\]
The square is important because potential energy is never negative in this spring reference.
A direct \(\sin\omega t\) dependence would make \(U\) negative during half the cycle, which does not match \(U=\frac{1}{2}kx^2\).
\( \textbf{Final answer:} \) \(U=\frac{1}{2}kA^2\sin^2\omega t\).
218. For a spring oscillator, \(K\) and \(U\) repeat their values twice during each complete displacement cycle. This happens because
ⓐ. the formulas contain squared quantities
ⓑ. the oscillator has two different amplitudes in each cycle
ⓒ. the time period of displacement is always zero
ⓓ. the force constant changes sign after every half cycle
Correct Answer: the formulas contain squared quantities
Explanation: In spring SHM, potential energy depends on \(x^2\), and kinetic energy depends on \(v^2\). Squaring removes the sign of displacement or velocity. Thus, the energy value at \(+x\) is the same as at \(-x\), and the energy value for velocity \(+v\) is the same as for \(-v\). During one full displacement cycle, these equal energy states occur twice. This is why \(K\) and \(U\) vary with twice the frequency of the displacement, while their sum remains constant.
219. A spring oscillator has total energy \(E\). At a certain position, \(U=\frac{3E}{4}\). The magnitude of displacement at that position is
ⓐ. \(\frac{A}{2}\)
ⓑ. \(\frac{\sqrt{3}A}{2}\)
ⓒ. \(\frac{A}{\sqrt{3}}\)
ⓓ. \(\frac{\sqrt{3}A}{4}\)
Correct Answer: \(\frac{\sqrt{3}A}{2}\)
Explanation: \( \textbf{Given:} \) \(U=\frac{3E}{4}\).
For a spring oscillator,
\[
U=\frac{1}{2}kx^2
\]
and
\[
E=\frac{1}{2}kA^2
\]
Take the ratio:
\[
\frac{U}{E}=\frac{\frac{1}{2}kx^2}{\frac{1}{2}kA^2}
\]
\[
\frac{U}{E}=\frac{x^2}{A^2}
\]
Given \(\frac{U}{E}=\frac{3}{4}\), so
\[
\frac{x^2}{A^2}=\frac{3}{4}
\]
\[
x^2=\frac{3A^2}{4}
\]
Taking magnitude,
\[
|x|=\frac{\sqrt{3}A}{2}
\]
The positive and negative positions have the same potential energy because the expression contains \(x^2\).
\( \textbf{Final answer:} \) \(|x|=\frac{\sqrt{3}A}{2}\).
220. A spring oscillator has amplitude \(A\). If the amplitude is changed to \(\frac{A}{2}\) while the same spring is used, the total energy becomes
ⓐ. \(\frac{E}{4}\)
ⓑ. \(\frac{E}{2}\)
ⓒ. \(2E\)
ⓓ. \(4E\)
Correct Answer: \(\frac{E}{4}\)
Explanation: The total energy of a spring oscillator is
\[
E=\frac{1}{2}kA^2
\]
For the same spring, \(k\) remains unchanged.
If the new amplitude is
\[
A'=\frac{A}{2}
\]
then the new energy is
\[
E'=\frac{1}{2}k(A')^2
\]
Substitute \(A'=\frac{A}{2}\):
\[
E'=\frac{1}{2}k\left(\frac{A}{2}\right)^2
\]
\[
E'=\frac{1}{2}k\frac{A^2}{4}
\]
\[
E'=\frac{1}{4}\left(\frac{1}{2}kA^2\right)
\]
\[
E'=\frac{E}{4}
\]
Energy follows the square of amplitude, not the first power of amplitude.
\( \textbf{Final answer:} \) \(E'=\frac{E}{4}\).
