Question

A sonometer is set on the floor of a lift. When the lift is at rest, the sonometer wire vibrates with fundamental frequency 256 Hz. When the lift goes up with acceleration $ a = \dfrac{{9g}}{{16}} $, the frequency of vibration of the same wire changes to
(A) 512 Hz
(B) 320 Hz
(C) 256 Hz
(D) 204 Hz

Answer 1

Hint: The frequency of the vibration of the sonometer is directly proportional to the square root acceleration experienced by the sonometer. The acceleration experienced in the second instance would be the sum of the acceleration of the lift and the acceleration due to gravity.

Formula used: In this solution we will be using the following formula;
 $ f \propto \sqrt a $ where $ f $ is the frequency of the vibration, $ a $ is the acceleration experienced by the string.

Complete step by step answer
The frequency of vibration of a sonometer is said to be found after the lift starts to accelerate upward.
In general, the frequency is directly proportional to the square root of the tension in the strings of the sonometer, which in turn is proportional to the weight of the string and in turn proportional to the acceleration experienced by the string. Hence, in summary
 $ f \propto \sqrt a $
 $ \Rightarrow f = k\sqrt a $ where $ f $ is the frequency of the vibration, $ a $ is the acceleration experienced by the string, and $ k $ is a proportionality constant.
Now when the elevator was at rest $ a = g $ where $ g $ is the acceleration due to gravity.
Hence, at rest
 $ 256 = k\sqrt g $
 $ \Rightarrow k = \dfrac{{256}}{{\sqrt g }} $
Now, at the time when lift is accelerating upwards at a rate of $ \dfrac{{9g}}{{16}} $ the total acceleration can be given as
 $ a = \dfrac{{9g}}{{16}} + g = \dfrac{{25g}}{{16}} $
Then the frequency would be given by
 $ {f_2} = k\sqrt {\dfrac{{25g}}{{16}}} $
Since $ k = \dfrac{{256}}{{\sqrt g }} $
Then we can write
 $ {f_2} = \dfrac{{256}}{{\sqrt g }}\sqrt {\dfrac{{25g}}{{16}}} $
This can be written as
 $ {f_2} = \dfrac{{256}}{{\sqrt g }}\sqrt {\dfrac{{25}}{{16}}} \sqrt g $
We can cancel $ \sqrt g $
Then
 $ {f_2} = 256\sqrt {\dfrac{{25}}{{16}}} = 256\left( {\dfrac{5}{4}} \right) $
 $ \Rightarrow {f_2} = 320Hz $

Hence, the correct answer is B.

Note
Alternatively, without finding the expression for the proportionality constant $ k $, we can simply compare as in since,
 $ {f_1} = k\sqrt g $ and
 $ {f_2} = k\sqrt {\dfrac{{25g}}{{16}}} $
Then by diving $ {f_2} $ by $ {f_1} $, we have
 $ \dfrac{{{f_2}}}{{{f_1}}} = k\sqrt {\dfrac{{25g}}{{16}}} \div k\sqrt g $
By converting the division into multiplication, and cancelling $ k $ we have
 $ \dfrac{{{f_2}}}{{{f_1}}} = \sqrt {\dfrac{{25g}}{{16}}} \times \dfrac{1}{{\sqrt g }} $
 $ \Rightarrow \dfrac{{{f_2}}}{{{f_1}}} = \dfrac{5}{4}\sqrt g \times \dfrac{1}{{\sqrt g }} $
Cancelling $ \sqrt g $ and multiplying by $ {f_1} $
 $ {f_2} = \dfrac{5}{4} \times {f_1} $
Hence, by inserting the values, we get
 $ \Rightarrow {f_2} = 320Hz $.