Question

The kinetic energy of the entire string at $t=0$ is matched correctly in the following option.

• A. Zero
• B. $\frac{a}{2}$
• C. $\frac{\pi^2 T a^2}{8L}$
• D. $a$
• E. $\frac{L}{4}$
• F. $\frac{\pi^2 T a^2}{4L}$

Answer 1

The kinetic energy of the entire string at t=0 for cases (I), (II), (III), and (IV) are 0, $\frac{\pi^2 T a^2}{4L}$, $\frac{\pi^2 T a^2}{L}$, and 0 respectively. Based on List-II, (I) matches with (P), (II) matches with (U), and (IV) matches with (P). The value for (III) is not present in List-II.

Answer 2

The kinetic energy of a small element of the string of length $dx$ at position $x$ is $dK = \frac{1}{2} dm v(x, t)^2 = \frac{1}{2} \mu (\frac{\partial y}{\partial t})^2 dx$, where $\mu$ is the linear mass density of the string.

The total kinetic energy of the string at time $t$ is $K(t) = \int_0^L \frac{1}{2} \mu \left(\frac{\partial y}{\partial t}\right)^2 dx$.

The wave speed on the string is $v_w = \sqrt{\frac{T}{\mu}}$. For a standing wave, the angular frequency $\omega$ and wave number $k$ are related by $\omega = k v_w = k \sqrt{\frac{T}{\mu}}$. Thus, $\mu = \frac{k^2 T}{\omega^2}$.

Let's calculate $K(0)$ for each case in List-I:

(I) $y = a \sin(\frac{\pi x}{L})\cos \omega t$. $\frac{\partial y}{\partial t} = -a \omega \sin(\frac{\pi x}{L})\sin \omega t$. At $t=0$, $\frac{\partial y}{\partial t}(x, 0) = -a \omega \sin(\frac{\pi x}{L})\sin(0) = 0$. $K(0) = \int_0^L \frac{1}{2} \mu (0)^2 dx = 0$. This matches with (P).

(II) $y = a \sin(\frac{\pi x}{L})\sin \omega t$. $\frac{\partial y}{\partial t} = a \omega \sin(\frac{\pi x}{L})\cos \omega t$. At $t=0$, $\frac{\partial y}{\partial t}(x, 0) = a \omega \sin(\frac{\pi x}{L})\cos(0) = a \omega \sin(\frac{\pi x}{L})$. $K(0) = \int_0^L \frac{1}{2} \mu (a \omega \sin(\frac{\pi x}{L}))^2 dx = \frac{1}{2} \mu a^2 \omega^2 \int_0^L \sin^2(\frac{\pi x}{L}) dx$. Using $\int_0^L \sin^2(\frac{n\pi x}{L}) dx = \frac{L}{2}$ for integer $n \ge 1$. $K(0) = \frac{1}{2} \mu a^2 \omega^2 \frac{L}{2} = \frac{1}{4} \mu a^2 \omega^2 L$. For this wave, the wave number is $k = \frac{\pi}{L}$. So $\mu = \frac{k^2 T}{\omega^2} = \frac{(\pi/L)^2 T}{\omega^2} = \frac{\pi^2 T}{\omega^2 L^2}$. $K(0) = \frac{1}{4} \left(\frac{\pi^2 T}{\omega^2 L^2}\right) a^2 \omega^2 L = \frac{\pi^2 T a^2}{4L}$. This matches with (U).

(III) $y = a \sin(\frac{2\pi x}{L})\sin \omega t$. $\frac{\partial y}{\partial t} = a \omega \sin(\frac{2\pi x}{L})\cos \omega t$. At $t=0$, $\frac{\partial y}{\partial t}(x, 0) = a \omega \sin(\frac{2\pi x}{L})\cos(0) = a \omega \sin(\frac{2\pi x}{L})$. $K(0) = \int_0^L \frac{1}{2} \mu (a \omega \sin(\frac{2\pi x}{L}))^2 dx = \frac{1}{2} \mu a^2 \omega^2 \int_0^L \sin^2(\frac{2\pi x}{L}) dx$. Using $\int_0^L \sin^2(\frac{n\pi x}{L}) dx = \frac{L}{2}$ for integer $n \ge 1$. $K(0) = \frac{1}{2} \mu a^2 \omega^2 \frac{L}{2} = \frac{1}{4} \mu a^2 \omega^2 L$. For this wave, the wave number is $k = \frac{2\pi}{L}$. So $\mu = \frac{k^2 T}{\omega^2} = \frac{(2\pi/L)^2 T}{\omega^2} = \frac{4\pi^2 T}{\omega^2 L^2}$. $K(0) = \frac{1}{4} \left(\frac{4\pi^2 T}{\omega^2 L^2}\right) a^2 \omega^2 L = \frac{\pi^2 T a^2}{L}$. This value is not present in List-II. Let's re-examine the problem and options. It is possible that the question intends to ask for the maximum kinetic energy or the total energy, or there might be an error in the options provided in List-II for case (III). Let's assume the question and options are correct and look for a potential misinterpretation or alternative calculation.

Let's check the potential energy at $t=0$ for case (III). $y = a \sin(\frac{2\pi x}{L})\sin \omega t$. $\frac{\partial y}{\partial x} = a \frac{2\pi}{L} \cos(\frac{2\pi x}{L})\sin \omega t$. At $t=0$, $\frac{\partial y}{\partial x}(x, 0) = a \frac{2\pi}{L} \cos(\frac{2\pi x}{L})\sin(0) = 0$. The total potential energy at $t=0$ is $U(0) = \int_0^L \frac{1}{2} T (\frac{\partial y}{\partial x}(x, 0))^2 dx = \int_0^L \frac{1}{2} T (0)^2 dx = 0$. For a standing wave, the total energy $E = K(t) + U(t)$ is constant. The kinetic energy is maximum when the potential energy is minimum (which is 0) and vice versa. For a standing wave of the form $y(x, t) = f(x) g(t)$, the kinetic energy is $K(t) = \int_0^L \frac{1}{2} \mu f(x)^2 (g'(t))^2 dx$ and potential energy is $U(t) = \int_0^L \frac{1}{2} T (f'(x))^2 g(t)^2 dx$. For case (III), $f(x) = a \sin(\frac{2\pi x}{L})$ and $g(t) = \sin \omega t$. $f'(x) = a \frac{2\pi}{L} \cos(\frac{2\pi x}{L})$ and $g'(t) = \omega \cos \omega t$. $K(t) = \int_0^L \frac{1}{2} \mu a^2 \sin^2(\frac{2\pi x}{L}) (\omega \cos \omega t)^2 dx = \frac{1}{2} \mu a^2 \omega^2 \cos^2 \omega t \int_0^L \sin^2(\frac{2\pi x}{L}) dx = \frac{1}{2} \mu a^2 \omega^2 \cos^2 \omega t \frac{L}{2} = \frac{1}{4} \mu a^2 \omega^2 L \cos^2 \omega t$. $U(t) = \int_0^L \frac{1}{2} T (a \frac{2\pi}{L} \cos(\frac{2\pi x}{L}))^2 (\sin \omega t)^2 dx = \frac{1}{2} T a^2 \frac{4\pi^2}{L^2} \sin^2 \omega t \int_0^L \cos^2(\frac{2\pi x}{L}) dx = \frac{1}{2} T a^2 \frac{4\pi^2}{L^2} \sin^2 \omega t \frac{L}{2} = \frac{\pi^2 T a^2}{L} \sin^2 \omega t$. Using $\mu = \frac{k^2 T}{\omega^2} = \frac{(2\pi/L)^2 T}{\omega^2} = \frac{4\pi^2 T}{\omega^2 L^2}$, we have $\mu \omega^2 = \frac{4\pi^2 T}{L^2}$. $K(t) = \frac{1}{4} (\frac{4\pi^2 T}{L^2}) a^2 L \cos^2 \omega t = \frac{\pi^2 T a^2}{L} \cos^2 \omega t$. $E = K(t) + U(t) = \frac{\pi^2 T a^2}{L} \cos^2 \omega t + \frac{\pi^2 T a^2}{L} \sin^2 \omega t = \frac{\pi^2 T a^2}{L} (\cos^2 \omega t + \sin^2 \omega t) = \frac{\pi^2 T a^2}{L}$. The total energy is indeed $\frac{\pi^2 T a^2}{L}$. At $t=0$, $K(0) = \frac{\pi^2 T a^2}{L} \cos^2(0) = \frac{\pi^2 T a^2}{L}$. $U(0) = \frac{\pi^2 T a^2}{L} \sin^2(0) = 0$. So, for case (III), $K(0) = \frac{\pi^2 T a^2}{L}$. This is still not in the list.

Let's check case (II) again using the total energy approach. $y = a \sin(\frac{\pi x}{L})\sin \omega t$. $k = \frac{\pi}{L}$. $K(t) = \frac{1}{4} \mu a^2 \omega^2 L \cos^2 \omega t$. $\mu \omega^2 = k^2 T = (\pi/L)^2 T = \frac{\pi^2 T}{L^2}$. $K(t) = \frac{1}{4} (\frac{\pi^2 T}{L^2}) a^2 L \cos^2 \omega t = \frac{\pi^2 T a^2}{4L} \cos^2 \omega t$. $U(t) = \int_0^L \frac{1}{2} T (a \frac{\pi}{L} \cos(\frac{\pi x}{L}))^2 (\sin \omega t)^2 dx = \frac{1}{2} T a^2 \frac{\pi^2}{L^2} \sin^2 \omega t \int_0^L \cos^2(\frac{\pi x}{L}) dx = \frac{1}{2} T a^2 \frac{\pi^2}{L^2} \sin^2 \omega t \frac{L}{2} = \frac{\pi^2 T a^2}{4L} \sin^2 \omega t$. $E = K(t) + U(t) = \frac{\pi^2 T a^2}{4L} \cos^2 \omega t + \frac{\pi^2 T a^2}{4L} \sin^2 \omega t = \frac{\pi^2 T a^2}{4L}$. At $t=0$, $K(0) = \frac{\pi^2 T a^2}{4L} \cos^2(0) = \frac{\pi^2 T a^2}{4L}$. $U(0) = \frac{\pi^2 T a^2}{4L} \sin^2(0) = 0$. So, for case (II), $K(0) = \frac{\pi^2 T a^2}{4L}$. This matches with (U).

Let's check case (I) using the total energy approach. $y = a \sin(\frac{\pi x}{L})\cos \omega t$. $k = \frac{\pi}{L}$. $K(t) = \frac{1}{4} \mu a^2 \omega^2 L \sin^2 \omega t = \frac{\pi^2 T a^2}{4L} \sin^2 \omega t$. $U(t) = \frac{\pi^2 T a^2}{4L} \cos^2 \omega t$. $E = K(t) + U(t) = \frac{\pi^2 T a^2}{4L}$. At $t=0$, $K(0) = \frac{\pi^2 T a^2}{4L} \sin^2(0) = 0$. $U(0) = \frac{\pi^2 T a^2}{4L} \cos^2(0) = \frac{\pi^2 T a^2}{4L}$. So, for case (I), $K(0) = 0$. This matches with (P).

Let's check case (IV) using the total energy approach. $y = a \sin(\frac{2\pi x}{L})\cos \omega t$. $k = \frac{2\pi}{L}$. $K(t) = \frac{1}{4} \mu a^2 \omega^2 L \sin^2 \omega t$. $\mu \omega^2 = k^2 T = (\frac{2\pi}{L})^2 T = \frac{4\pi^2 T}{L^2}$. $K(t) = \frac{1}{4} (\frac{4\pi^2 T}{L^2}) a^2 L \sin^2 \omega t = \frac{\pi^2 T a^2}{L} \sin^2 \omega t$. $U(t) = \int_0^L \frac{1}{2} T (a \frac{2\pi}{L} \cos(\frac{2\pi x}{L}))^2 (\cos \omega t)^2 dx = \frac{1}{2} T a^2 \frac{4\pi^2}{L^2} \cos^2 \omega t \int_0^L \cos^2(\frac{2\pi x}{L}) dx = \frac{1}{2} T a^2 \frac{4\pi^2}{L^2} \cos^2 \omega t \frac{L}{2} = \frac{\pi^2 T a^2}{L} \cos^2 \omega t$. $E = K(t) + U(t) = \frac{\pi^2 T a^2}{L} \sin^2 \omega t + \frac{\pi^2 T a^2}{L} \cos^2 \omega t = \frac{\pi^2 T a^2}{L}$. At $t=0$, $K(0) = \frac{\pi^2 T a^2}{L} \sin^2(0) = 0$. $U(0) = \frac{\pi^2 T a^2}{L} \cos^2(0) = \frac{\pi^2 T a^2}{L}$. So, for case (IV), $K(0) = 0$. This matches with (P).

Summary of matches based on $K(0)$: (I) - (P) (II) - (U) (III) - $\frac{\pi^2 T a^2}{L}$ (not in list) (IV) - (P)

Let's reconsider the options in List-II. Perhaps (R) $\frac{\pi^2 T a^2}{8L}$ and (U) $\frac{\pi^2 T a^2}{4L}$ are related to the total energy or maximum kinetic energy. Maximum kinetic energy occurs when $\cos^2 \omega t = 1$ (for sine time dependence) or $\sin^2 \omega t = 1$ (for cosine time dependence). Maximum kinetic energy for (I) is $\frac{\pi^2 T a^2}{4L}$. Maximum kinetic energy for (II) is $\frac{\pi^2 T a^2}{4L}$. Maximum kinetic energy for (III) is $\frac{\pi^2 T a^2}{L}$. Maximum kinetic energy for (IV) is $\frac{\pi^2 T a^2}{L}$.

The total energy for (I) and (II) is $\frac{\pi^2 T a^2}{4L}$. This matches (U). The total energy for (III) and (IV) is $\frac{\pi^2 T a^2}{L}$. This is not in the list.

Let's check if any option in List-II represents the maximum displacement (amplitude) or position of nodes/antinodes. (Q) $\frac{a}{2}$ is half the amplitude. (S) $a$ is the amplitude. (T) $\frac{L}{4}$ is a position along the string. For the fundamental mode ($n=1$, $k=\pi/L$), $x=L/4$ is a point between a node and an antinode. For the second harmonic ($n=2$, $k=2\pi/L$), $x=L/4$ is an antinode position.

Let's assume there is a typo in the question or the options, specifically for case (III). If the question was asking for something else related to case (III), or if the formula for (R) or (U) was different.

Given the provided solution structure asks for the correct option number, this question is likely a single-choice question where an option lists the correct matchings for (I), (II), (III), and (IV). Since the options are not provided, we cannot definitively choose the correct option. However, based on our calculations: (I) matches with (P) (II) matches with (U) (IV) matches with (P)

Let's assume there is a match for (III) in the options. Let's recheck the calculation for (III) one more time. $K(0) = \frac{1}{4} \mu a^2 \omega^2 L$. $\mu = \frac{k^2 T}{\omega^2}$ where $k = \frac{2\pi}{L}$. $K(0) = \frac{1}{4} \frac{(2\pi/L)^2 T}{\omega^2} a^2 \omega^2 L = \frac{1}{4} \frac{4\pi^2}{L^2} T a^2 L = \frac{\pi^2 T a^2}{L}$.

Let's assume there is a typo in the wave equation for (III) or (IV). If (III) was $y = a \sin(\frac{\pi x}{L})\sin \omega t$, it would be the same as (II), giving $K(0) = \frac{\pi^2 T a^2}{4L}$ (U). If (III) was $y = a \sin(\frac{\pi x}{L})\cos \omega t$, it would be the same as (I), giving $K(0) = 0$ (P). If the wave number in (III) was different, e.g., $k=\pi/(2L)$, it would not be a standing wave on a string fixed at both ends of length L.

Let's assume there is a typo in option (R) or (U). If (U) was $\frac{\pi^2 T a^2}{L}$, then (III) would match with (U). But (II) matches with (U) as given. If (R) was $\frac{\pi^2 T a^2}{L}$, then (III) would match with (R). If (R) was $\frac{\pi^2 T a^2}{4L}$, then (II) would match with (R) and (U) would be unused or incorrect.

Let's assume the question asks for something else for (III). What if it asks for the kinetic energy at a specific time? Or the potential energy at t=0? For (III), $U(0) = 0$ (P). For (I), $U(0) = \frac{\pi^2 T a^2}{4L}$ (U). For (II), $U(0) = 0$ (P). For (IV), $U(0) = \frac{\pi^2 T a^2}{L}$ (not in list).

Let's assume there is a typo in the question and for (III) the wave number is $\pi/L$ instead of $2\pi/L$. Then (III) is identical to (II) and $K(0)=U$. Let's assume there is a typo in option (R). If (R) was $\frac{\pi^2 T a^2}{L}$, then (III) matches with (R).

Given that the question asks for a correct option from a set of matchings, and our derived match for (III) is not in the list, there is likely an error in the question or the provided options in List-II. However, if we must choose the best possible answer based on the options provided, and assuming there is a single correct option among the choices (which are not given), we would look for an option that contains (I)-(P), (II)-(U), and (IV)-(P). The matching for (III) in that option would likely correspond to one of the values in List-II, possibly indicating a typo in the question or the list.

Assuming the question intended to have a valid match for (III), and given the options (R) and (U), the value $\frac{\pi^2 T a^2}{L}$ for (III) is $4 \times \frac{\pi^2 T a^2}{4L} = 4 \times (U)$. It is also $8 \times \frac{\pi^2 T a^2}{8L} = 8 \times (R)$. Neither (R) nor (U) match (III).

Let's assume, hypothetically, that one of the options for the full matching was (I)-(P), (II)-(U), (III)-(R), (IV)-(P). This would imply that $K(0)$ for (III) is (R) $\frac{\pi^2 T a^2}{8L}$. This contradicts our calculation.

Let's assume, hypothetically, that one of the options for the full matching was (I)-(P), (II)-(U), (III)-(U), (IV)-(P). This would imply that $K(0)$ for (III) is (U) $\frac{\pi^2 T a^2}{4L}$. This also contradicts our calculation.

Given the strong consistency of our calculation for (I), (II), and (IV) matching (P) and (U), and the inconsistency for (III), it is most likely that there is an error in the problem statement regarding case (III) or the options in List-II.