Question

Standing waves are produced in $$10m$$ long stretched strings. If the string vibrates in 5 segments and wave velocity is $$20m{s}^{-1}$$ , then its frequency will be
A. $$5Hz$$
B. $$2Hz$$
C. $$10Hz$$
D. $$12Hz$$

Answer 1

Hint: To answer the question, we will build a simple diagram based on the question. The entire length of the string is $$10m$$ , and there are five segments, so we will compute the length of one segment and determine the value of $\lambda$ , and then calculate the frequency $(\nu )$ using this $\lambda$ .

Complete answer:
Before we go into the question, let's have a look at what a standing wave is. The combination of two waves flowing in opposite directions, each with the same amplitude and frequency, is known as a standing wave.
Now, let us come to the question;
The wavelength of a stretched string's fundamental vibrational mode is twice the length of the string.

Because the string produces standing waves and vibrates in five parts, it can be shown as
$$\therefore 5{\displaystyle \frac{\lambda }{2}}=10$$
Therefore, from here we will find value of $\lambda$
$$\Rightarrow \lambda =4m$$
The wave's velocity, $$v$$ , is given to us in the question as $v=20m{s}^{-1}$
Hence, the frequency will be $$\nu ={\displaystyle \frac{v}{\lambda }}={\displaystyle \frac{20}{4}}=5s^{-1}=5Hz$$
Therefore, the frequency is $$5Hz$$
The correct option is: (A) $$5Hz$$

Note:
It's important to note that standing waves don't just appear out of nowhere. They call for energy to be delivered into a system at a specific frequency. That is, when a system's driving frequency is identical to its natural frequency. Resonance is the term for this situation. Standing waves are invariably linked to resonance.