Physics Q101 – AP-EAMCET 2026

Two identical strings each of length

0.750 m

are each tuned exactly to

440 Hz

. The tension in one of the strings is then increased by

1.0 %

. If they are now struck, then the beat frequency between the fundamentals of the two strings is nearly

Clue 1 of 3: Fundamental Frequency of a String

The fundamental frequency of vibration of a stretched string is determined by its length, tension, and linear mass density.

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Step 1: Establish the relationship between frequency and tension✦ Active

The fundamental frequency $f$ of a vibrating string is given by the formula:

f = \frac{1}{2 L} \sqrt{\frac{T}{μ}}

where $L$ is the length, $T$ is the tension, and $μ$ is the linear mass density. For two identical strings, $L$ and $μ$ are constant. Therefore, the frequency is directly proportional to the square root of the tension:

f \propto \sqrt{T}

💡 Teacher's Secret Hint

Remember that for a given string, length and mass density remain constant unless specified. This simplifies the relationship to just tension.

Step 2: Calculate the new frequency of the altered string○ Expand

Let the initial frequency of both strings be $f_{0} = 440 Hz$ and the initial tension be $T_{0}$ . The tension in one string is increased by $1.0 %$ . So, the new tension $T^{'}$ is $T_{0} + 0.01 T_{0} = 1.01 T_{0}$ . The frequency of the other string remains $f_{0}$ .

The new frequency $f^{'}$ of the altered string can be found using the proportionality derived in Step 1:

\frac{f^{'}}{f_{0}} = \frac{\sqrt{T^{'}}}{\sqrt{T_{0}}} = \sqrt{\frac{T^{'}}{T_{0}}}

Substitute $T^{'} = 1.01 T_{0}$ into the equation:

\frac{f^{'}}{f_{0}} = \sqrt{\frac{1.01 T_{0}}{T_{0}}} = \sqrt{1.01}

Using the binomial approximation $(1 + x)^{n} \approx 1 + n x$ for small $x$ , where $x = 0.01$ and $n = 1 / 2$ :

\sqrt{1.01} = (1 + 0.01)^{1 / 2} \approx 1 + \frac{1}{2} (0.01) = 1 + 0.005 = 1.005

So, the new frequency is approximately $f^{'} \approx 1.005 f_{0}$ . Substituting the value of $f_{0}$ :

f^{'} \approx 1.005 \times 440 Hz = 442.2 Hz

💡 Teacher's Secret Hint

The binomial approximation is very useful for small percentage changes. A more precise calculation of $\sqrt{1.01}$ yields $\approx 1.0049875$ .

Step 3: Calculate the beat frequency○ Expand

The beat frequency $f_{b e a t}$ is the absolute difference between the frequencies of the two strings:

f_{b e a t} = | f^{'} - f_{0} |

Substitute the approximate value of $f^{'}$ :

f_{b e a t} = | 442.2 Hz - 440 Hz | = 2.2 Hz

Alternatively, using $f^{'} \approx 1.005 f_{0}$ :

f_{b e a t} = | 1.005 f_{0} - f_{0} | = | (1.005 - 1) f_{0} | = 0.005 f_{0}

f_{b e a t} = 0.005 \times 440 Hz = 2.2 Hz

💡 Teacher's Secret Hint

Remember that beat frequency is always a positive value, hence the absolute difference. The term 'nearly' in the question indicates that an approximate answer or the closest integer/simple fraction is expected.

Step 4: Select the closest option○ Expand

The calculated beat frequency is approximately $2.2 Hz$ . Among the given options:

1. $1 Hz$

2. $2 Hz$

3. $3 Hz$

4. No beats

The value $2.2 Hz$ is closest to $2 Hz$ .

💡 Teacher's Secret Hint

Always check the precision required by the options. Sometimes, a more precise calculation for $\sqrt{1.01}$ might lead to $2.19 Hz$ , which still rounds to $2 Hz$ when considering options are integers.

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Physics Question 101 – AP-EAMCET 2026

Clue 1 of 3: Fundamental Frequency of a String

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