Question

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is $8.0N$. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to
(A) $16.6$ cm
(B) $20.0$ cm
(C) $10.0$ cm
(D) $33.3$ cm

Answer 1

Hint
To find the separation between successive nodes on the string, we need to calculate the velocity of the wave on the string then, we have to find out the wavelength. The half of the wavelength is equal to the separation between the nodes.
$\Rightarrow v = \sqrt {\dfrac{T}{m}} $$\sqrt l $
$\Rightarrow \lambda = \dfrac{v}{n}$
Where, the wavelength of the string is λ, the velocity be v, the length of the string be l, tension in the string be T.

Complete step by step answer
To calculate the velocity of the wave in the string, we need to use the formula,
$\Rightarrow v = \sqrt {\dfrac{T}{m}} $$\sqrt l $
Now, we know the tension in the string as 8 N and the mass is 5g, the length of the string is 1m.
$\Rightarrow v = \sqrt {\dfrac{8}{5} \times 1000} $
$\Rightarrow v = 40m/s$
Now, the wavelength of the string is given by,
$\Rightarrow \lambda = \dfrac{v}{n}$
Where v is the velocity of the string and n is the frequency.
Then,
$\Rightarrow \lambda = \dfrac{{40}}{{100}}m$
Now, we have to divide the wavelength by 2, to get the separation between successive nodes of the string.
$\Rightarrow \lambda = \dfrac{{\dfrac{{40}}{{100}}}}{2}$
Then,
$\Rightarrow \lambda = 0.2m \\
   \Rightarrow \lambda = 20cm \\ $
Thus, we get the separation between successive nodes of the string as 20cm.
Hence, the correct answer is option (B).

Note
A standing wave on a string fixed on the both sides is the superimposition of the two waves of equal amplitude and the frequency but travels in the opposite direction. When the two opposite waves meet at the midpoint in equilibrium condition, then the number of the nodes and the number of the antinodes will be equal.