A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the st

Q: Why do we use the sine term for the spatial part $kx$?

Since the string is clamped at $x=0$, the displacement $Y$ must be zero at $x=0$ for all time $t$. The sine function naturally satisfies this boundary condition ($\sin(0) = 0$).

Q: How did we know it forms 4 loops?

The problem explicitly states the string is vibrating in its '4th harmonic'. For a string fixed at both ends, the $n^{\text{th}}$ harmonic corresponds to exactly $n$ loops.

Q: Is $0.157$ always $\frac{\pi}{20}$?

In the context of physics problems where $\pi \approx 3.14$, dividing by 20 gives $0.157$. Recognizing these numerical patterns saves crucial time during competitive exams.

Q: What does the cosine term represent?

The $\cos(200\pi t)$ term represents the temporal oscillation. It tells us that all particles on the string oscillate in simple harmonic motion with an angular frequency $\omega = 200\pi \text{ rad/s}$.

Q: Why is the length $L = 2\lambda$?

Each loop of a standing wave corresponds to half a wavelength ($\frac{\lambda}{2}$). Since the 4th harmonic has 4 loops, the total length is $4 \times \frac{\lambda}{2} = 2\lambda$.

Question

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y = 0.3 \sin(0.157x) \cos(200\pi t)$. The length of the string is (All quantities are in SI units)

Options:

(A) 60 m

(B) 40 m

(C) 80 m

(D) 20 m

Sigma Percentile · Accepted Answer

The problem of a vibrating string is a classic in physics, beautifully marrying geometry with wave mechanics. When a string is clamped at both ends, it cannot vibrate just any way it wants. It is forced to form specific patterns called **standing waves** or **stationary waves**.

### Visualizing the 4th Harmonic

Imagine you are holding a string that is fixed at both ends. If you pluck it just right, it vibrates in its **4th harmonic**. What does this mean physically? It means the string naturally divides itself into exactly 4 distinct vibrating segments, or "loops".

Every single loop in a standing wave has a length of exactly $\frac{\lambda}{2}$, where $\lambda$ is the wavelength. Since our string has 4 loops, the total length $L$ of the string must be:

$L = 4 	imes \frac{\lambda}{2} = 2\lambda$

This geometric constraint is our master key. If we can find the wavelength $\lambda$, we can instantly find the length of the string.

### The Master Equation

The problem provides us with the mathematical heartbeat of this string:

$Y = 0.3 \sin(0.157x) \cos(200\pi t)$

I know this looks like a dense alphabet soup, but let's take a breath and compare it to the standard equation of a standing wave:

$Y = A \sin(kx) \cos(\omega t)$

By simply matching the terms, we can extract the physical properties of the wave. The spatial part (the part with $x$) gives us the **wave number**, $k$.

$k = 0.157 	ext{ m}^{-1}$

### The Pi Approximation Trick

Now, we need to connect the wave number $k$ to the wavelength $\lambda$. The fundamental relationship is:

$k = \frac{2\pi}{\lambda}$

So, we have:

$\frac{2\pi}{\lambda} = 0.157$

Here is where a lot of students get stuck doing messy decimal division. But there is a catch here—a beautiful numerical trick that is a favorite in JEE and NEET exams. Notice that $\pi \approx 3.1415$. If you divide $\pi$ by 20, you get exactly $0.157$!

Let's substitute this approximation into our equation:

$\frac{2\pi}{\lambda} = \frac{\pi}{20}$

The $\pi$ on both sides elegantly cancels out, leaving us with a simple cross-multiplication:

$\lambda = 2 	imes 20 = 40 	ext{ m}$

### Final Calculation

We are in the endgame now. Remember that value of $L$ we found earlier? Let's bring that back. We established that for the 4th harmonic, the length of the string is twice the wavelength.

$L = 2\lambda$

Substitute our newly found wavelength:

$L = 2(40)$

$L = 80 	ext{ m}$

And there we have it! The total length of the string is exactly 80 meters. By combining physical visualization with a clever mathematical approximation, we cracked the code of the standing wave.

Animated Solution for Physics - Waves: A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is $Y = 0.3 sin (0.157 x) cos (200 π t)$ . The length of the string is (All quantities are in SI units)

Select Answer:

Visualized Solution

$4^{th}$ Harmonic

Standard Equation of Standing Wave

Extracting Wave Number

Calculating Wavelength

Finding the Length

Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Visualizing the 4th Harmonic

The Master Equation

The Pi Approximation Trick

Final Calculation

Similar Questions

Length of a string tied to two rigid supports is $40 cm$ . Maximum length (wavelength in cm) of a stationary wave produced on it, is

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y (x, t) = (0.01 m) [sin (62.8 m^{- 1}) x] cos [(628 s^{- 1}) t]$ . Assuming $π = 3.14$ , the correct statement(s) is (are)

A $20 cm$ long string, having a mass of $1.0 g$ , is fixed at both the ends. The tension in the string is $0.5 N$ . The string is set into vibration using an external vibrator of frequency $100 Hz$ . Find the separation (in cm) between the successive nodes on the string.

A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} kgm^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are : $420 Hz$ and $490 Hz$ . Then, $L$ in metres is