Question

A stretched string is fixed at both ends, 77.7 cm apart. If the density of the string is 0.014 g/cm and its tension is 600 N, what is the wavelength of the first harmonic?

Answer 1

Answer:

The wavelength of the first harmonic is 155.4 cm.

Explanation:

Given that,

The separation between a stretched string that is fixed at both ends, l = 77.7 cm

The density of the string, $d=0.014\ g/cm^3$

Tension in the string, T = 600 N

We need to find the wavelength of the first harmonic. The wavelength of the nth harmonic on the string that is fixed at both ends is given by :

$\lambda=\dfrac{2l}{n}$

Here, n = 1

$\lambda=2l$

$\lambda=2\times 77.7$

$\lambda=155.4\ cm$

So, the wavelength of the first harmonic is 155.4 cm. Hence, this is the required solution.                                                              

Answer 2

The wavelength of the first harmonic will be "155.4 cm".

Wavelength and Density

According to the question,

Separation between stretched string, I = 77.7 cm

String's density, d = 0.014 g/cm³

String's tension, T = 600 N

We know the relation,

Wavelength, λ = $\frac{2l}{n}$

here, n = 1

then, λ = 2l

By substituting the values,

           = 2 × 77.7

           = 155.4 cm

Thus the above answer is appropriate.  

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