Question

The string of a sonometer is divided into two parts using a wedge. Total length of string is 1m and two parts differ by 2mm. when sounded together they produce 2Hz beats. The frequencies of two parts are
(A) 501 Hz, 503 Hz
(B) 501 Hz, 499 Hz
(C) 499 Hz, 497 Hz
(D) 497Hz, 495 Hz

Answer 1

Hint: The frequency can be given to be the velocity divided by the length of the parts of the string. Use simultaneous equations to find the length of the two parts of the string from the information given.

Formula used : In this solution we will be using the following formulae;
 $ f = \dfrac{v}{l} $ where $ f $ is the frequency of the waves on a string due to vibration of the string, $ v $ is the frequency of the wave, and $ l $ is the length of the string.
 $ \Delta f = {f_1} - {f_2} $ where $ \Delta f $ is the beat of two waves, $ {f_1} $ is the frequency of one of the wave and $ {f_2} $ is the frequency of the other.

Complete step by step answer:
As told, the total length of the string is 1 m. hence, we can write that
 $ x + y = 100cm $ (since 1 m equal 100 cm)
And the two length is said to differ by 2 mm. hence,
 $ y - x = 0.2cm $
 $ \Rightarrow y = 0.2 + x $
Inserting this into above equation, we get
 $ x + \left( {0.2 + x} \right) = 100 $
By calculating for $ x $ , we get
 $ x = \dfrac{{99.8}}{2} = 49.9cm $
Hence, $ y = 0.2 + x = 0.2 + 49.9 = 50.1cm $
The frequency of the vibration can be given as
 $ f = \dfrac{v}{l} $ where $ v $ is the frequency of the wave, and $ l $ is the length of the string.
the frequencies of x part and y part are hence
 $ {f_x} = \dfrac{v}{{0.501}} $ and $ {f_y} = \dfrac{v}{{0.499}} $ .
Now the beat can be given as
 $ \Delta f = {f_1} - {f_2} $ where $ \Delta f $ is the beat of two waves, $ {f_1} $ is the frequency of one of the wave and $ {f_2} $ is the frequency of the other.
Hence,
 $ \dfrac{v}{{0.499}} - \dfrac{v}{{0.501}} = 2 $
 $ \Rightarrow \dfrac{{0.501v - 0.499v}}{{0.499 \times 0.501}} = 2 $
Calculating for the velocity, we get
 $ v = \dfrac{{2 \times 0.499 \times 0.501}}{{0.002}} = 254.999m/s $
Hence, the frequency of x and y part of the string can be calculated as
 $ {f_y} = \dfrac{{254.999}}{{0.499}} = 501Hz $ and
 $ {f_x} = \dfrac{{254.999}}{{0.501}} = 499Hz $
Hence, the correct option is B.

Note:
Alternatively, you can calculate the frequency of x part from the knowledge of the beat, as in
 $ \Delta f = {f_y} - {f_x} $
 $ \Rightarrow {f_x} = {f_y} - \Delta f $
Then by inserting the known values
 $ \Rightarrow {f_x} = 501 - 2 = 499Hz $
Hence, the correct option B.