So the question goes like this

A string 1 m long is fixed at one end. The other end is moved up and down with frequency 15Hz. Due to this , a stationary wave with four complete loops , gets produced on the string. Find the speed of the progressive wave which produces the stationary wave [ Hint: Remember that the moving end is an antinode.]

basically any other text book homework question but litrary every answer i can find on internet is using the relation $$ \lambda = \frac{L}{2} $$ ($ \lambda $ is wavelength of standing wave formed)

but it seems wrong as the question clearly states one of the ends is an antinode which means $\lambda$ should be $\frac{4}{7}$ times the length of the spring, Am i missing something here or is litrary every website copying other answers without verifying them.


1 Answer 1


The wavelength is not related by either of those two. It's $\lambda = \frac{2*L}{n}$ where $n$ is the number of loops. Since it's 4 loops PLUS the half loop required to generate the string with an antinode, $\frac{2*1}{4.5} = .\overline{4}\text{meters}$

The speed of this wave is then $v = \lambda f = .\overline{4}\text{m}*15\text{Hz} = 6.\overline{6}\frac{m}{s}$ I'm not sure where you got $4/7$ from

  • $\begingroup$ I just drew the wave and measured it to get the 4/7 value, turns out i confused nodes with loops.. $\endgroup$ Dec 21, 2022 at 20:08
  • $\begingroup$ Nodes occur at the ends of loops, antinodes occur in the middle of loops. If the standing wave has 4.5 loops, it's got 4.5 nodes. $\endgroup$
    – Obliv
    Dec 21, 2022 at 20:36
  • $\begingroup$ Yeah i just read loops but thought of them as nodes for some reason, I dont undersatnd the next line it how could 4.5 loops have 4.5 nodes ig you don't count the fixed end as a node? image $\endgroup$ Dec 22, 2022 at 9:41
  • $\begingroup$ @MandoBites You DO count the fixed end as a node. For a complete loop, there are nodes at either end of the loop, correct? If you have 4 loops, there are nodes at either end of each loop, but you don't count the same node twice. Therefore, 4 loops means 4 unique nodes. Since you need an antinode to generate the standing wave, you have a half of a loop on top of this. A half loop is counted as .5 if a full loop is 1. $\endgroup$
    – Obliv
    Dec 22, 2022 at 16:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.