Question

A string 105 cm long is fixed at one end.the other end of string is moved up and down with frequency 15Hz A stationary wave produced in the string consist of 3 loops calculate the speed of progressive waves which have produced the stationary wave in the string​

Answer 1

Answer:

There are 5 loops in 10 metre.

Length of one loop =2m

Length of two loops =2×2=4m

Now, λ=4m;v=20m/sec

From formula, v=vλ.⇒20=v×4⇒v=5Hz

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Answer verified

Answer 2

Step by step explanation:

Given: String of $105cm$ and Frequency is $15Hz$.

To find: The speed of progressive waves.

For calculation of wavelength or 'λ',

We know that $l=105cm=3$ loops,

⇒ $l=3\frac{λ}{2}$

⇒ $λ = \frac{2}{3} l$

⇒ $\frac{2}{3}$ × $105$

⇒ $70 cm=0.70 m$

For calculation of speed,

We know that $v$ $= n$ × λ

⇒ $v=15$ × $0.70$

⇒ $10.50 m/s$

The speed of the progressive waves is $10.50 Hz$ $m/s$.