A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant fre

Q: Why do we assume the frequencies are consecutive harmonics?

The problem explicitly states 'There are no other resonant frequencies between these two.' This implies they must be adjacent harmonics, meaning if one is the $n^{\text{th}}$ harmonic, the next must be the $(n+1)^{\text{th}}$ harmonic.

Q: What exactly is the lowest resonant frequency?

The lowest resonant frequency is the fundamental frequency, which corresponds to the first harmonic ($n=1$). It is the lowest frequency at which a standing wave can be formed on the string.

Q: Why did we divide the two equations?

Dividing the equations is a mathematical trick to eliminate the unknown variables $v$ (wave speed) and $L$ (length). This leaves us with an equation containing only $n$, allowing us to easily solve for the harmonic number.

Q: Could we have found the wave speed $v$ from this data?

Yes! Once we know $f_1 = 105 \text{ Hz}$ and $L = 0.75 \text{ m}$, we can use the fundamental frequency formula $f_1 = \frac{v}{2L}$ to find $v = 2 \times 0.75 \times 105 = 157.5 \text{ m/s}$.

Q: Does the length $L=75.0 \text{ cm}$ affect the calculation of $n$?

No, the length $L$ cancels out when we take the ratio of the two frequencies. However, it would be absolutely necessary if we wanted to calculate the wave speed $v$ or the tension in the string.

Question

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Options:

(A) 105 Hz

(B) 1.05 Hz

(C) 1050 Hz

(D) 10.5 Hz

Sigma Percentile · Accepted Answer

### The Setup: A String Fixed at Both Ends

Imagine a string stretched tightly between two rigid walls. When plucked, it vibrates, creating standing waves. These standing waves can only exist at specific, discrete frequencies known as **resonant frequencies** or **harmonics**.

The problem tells us that the string has a length of $L = 75.0 	ext{ cm}$ and vibrates at two consecutive resonant frequencies: $315 	ext{ Hz}$ and $420 	ext{ Hz}$. The phrase "no other resonant frequencies between these two" is the golden key here—it tells us these are adjacent harmonics.

### The Master Equation: Resonant Frequencies

For a string fixed at both ends, the resonant frequency of the $n^{	ext{th}}$ harmonic is governed by the beautiful relation:

$$f_n = \frac{nv}{2L}$$

Here, $v$ is the speed of the transverse wave on the string, and $n$ is an integer ($1, 2, 3, \dots$) representing the harmonic number. The lowest possible frequency occurs when $n=1$, which is called the **fundamental frequency** ($f_1$). Notice that any higher harmonic is simply an integer multiple of the fundamental frequency: $f_n = n f_1$.

### The Consecutive Harmonics

Since $315 	ext{ Hz}$ and $420 	ext{ Hz}$ are consecutive, let's assign them to the $n^{	ext{th}}$ and $(n+1)^{	ext{th}}$ harmonics respectively. We can write out our two equations:

$$f_n = \frac{nv}{2L} = 315 	ext{ Hz}$$

$$f_{n+1} = \frac{(n+1)v}{2L} = 420 	ext{ Hz}$$

We have a system of equations, but we don't know $v$ or $n$. However, by dividing the two equations, the term $\frac{v}{2L}$ elegantly cancels out, leaving us with a simple ratio:

$$\frac{n}{n+1} = \frac{315}{420}$$

Simplifying the fraction on the right side by dividing the numerator and denominator by $105$, we get:

$$\frac{n}{n+1} = \frac{3}{4}$$

Cross-multiplying yields $4n = 3n + 3$, which immediately gives us $n = 3$. This means $315 	ext{ Hz}$ is the 3rd harmonic, and $420 	ext{ Hz}$ is the 4th harmonic.

### The Final Calculation: Uncovering the Fundamental

Now that we know $n=3$, we can easily find the lowest resonant frequency, which is the fundamental frequency $f_1$. Since the $n^{	ext{th}}$ harmonic is $n$ times the fundamental frequency, we have:

$$f_n = n f_1$$

Substituting our known values:

$$315 = 3 f_1$$

$$f_1 = \frac{315}{3} = 105 	ext{ Hz}$$

And there we have it! The lowest resonant frequency for this string is **$105 	ext{ Hz}$**. The beauty of this problem lies in how the ratio of consecutive frequencies directly reveals the harmonic number, bypassing the need to calculate the wave speed entirely.

Animated Solution for Physics - Waves: A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Select Answer:

Visualized Solution

$L = 75.0 cm$

$f_{n} = \frac{n v}{2 L}$

$Consecutive Frequencies$

$Equations$

$Dividing Equations$

$Solving for n$

$Lowest Resonant Frequency$

$f_{1} = 105 Hz$

$Extensions$

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Setup

The Master Equation

The Consecutive Harmonics

The Final Calculation

Similar Questions

A string of length $1 m$ and mass $2 \times 1 0^{- 5} kg$ is under tension $T$ . when the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is _______ Newton.

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

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by $100 Hz$ than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency $512 Hz$ . If the tube is opened at both ends the fundamental frequency that can be excited is (in Hz)

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.

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 string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to

Animated Solution for Physics - Waves: A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=75.0 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fn​=2Lnv​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consecutive Frequencies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dividing Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lowest Resonant Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f1​=105 Hz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extensions