Question

TWO IDENTICAL STRAIGHT WIRES ARE STRETCHED SO AS TO PRODUCE 6 BEATS/S WHEN VIBRATING SIMULTANEOUSLY. ON CHANGING THE TENSION IN ONE OF THEM , BEAT FREQUENCY REMAINS SAME. DENOTING BY T1 AND T2 , THE HIGHER AND THE LOWER INITIAL TENSION IN THE STRING , THEN IT COULD BE SAID THAT WHILE MAKING THE ABOVE CHANGES IN THE TENSION :
1. T2 WAS DECREASED
2. T2 WAS INCREASED
3. T1 WAS INCREASED
4. NONE OF THE ABOVE

Answer 1

Answer on Question #42021

Physics - Mechanics | Kinematics | Dynamics

Question:

TWO IDENTICAL STRAIGHT WIRES ARE STRETCHED SO AS TO PRODUCE 6 BEATS/S WHEN VIBRATING SIMULTANEOUSLY. ON CHANGING THE TENSION IN ONE OF THEM, BEAT FREQUENCY REMAINS SAME. DENOTING BY T1 AND T2, THE HIGHER AND THE LOWER INITIAL TENSION IN THE STRING, THEN IT COULD BE SAID THAT WHILE MAKING THE ABOVE CHANGES IN THE TENSION:

1. T2 WAS DECREASED

2. T2 WAS INCREASED

3. T1 WAS INCREASED

4. NONE OF THE ABOVE

Solution:

Frequency of the oscillations of a wire is determined by the following formula:

$$f = \frac {1}{2 L} \sqrt {\frac {T}{\rho S}},$$

where $L$ is a length, $\rho$ is a wire density and $S$ is an area of the wire cross-section. Thus, if $T_{1} > T_{2}$,

$$f _ {1} \sim \sqrt {T _ {1}} > f _ {2} \sim \sqrt {T _ {2}}.$$

The beat frequency is $\Delta f = |f_1 - f_2| = f_1 - f_2$. If it is not changed, it means that the new beat frequency equals $\Delta f = |f_1' - f_2| = f_2 - f_1'$. So, we can conclude, that $f_1' < f_1$ and $T_1$ was decreased. The answer is 4. None of the above.

Answer:

4. None of the above.

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