Question

A string, fixed at both ends, vibrates in a resonant node with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. find the length of the string.

• A. 18 cm
• B. 8 cm
• C. 10 cm
• D. cannot be determined

Answer 1

The correct option is B

8 cm

In the first case the distance between cause cu tine nodes is 2 cm,

Which implies $\frac{\lambda }{2}=2cm$ cm or $\lambda =4cm$

If v is the velocity of wave on the string

Frequency,$f=f=\frac{v}{\lambda }=\frac{v}{4}$- - - - -- (1)

When the distance between the nodes become 1.6 cm

$\frac{{\lambda }^{\prime }}{2}=1.6cm$ $\Rightarrow {\lambda }^{\prime }=3.2cm$

Frequency,$f^{\prime }=\frac{v}{{\lambda }^{\prime }}=\frac{v}{3.2}$ - - - - - (2)

Also for some n,$f=\frac{nv}{2L}$

$\Rightarrow \frac{v}{4}=\frac{nv}{2L}\left[from\right(1\left)\right]$

$n=\frac{L}{2}$ - - - - - (3)

then for some $n+1,f^{\prime }=\frac{(n+1)v}{2L}$

$\Rightarrow \frac{v}{3.2}=\frac{(n+1)v}{2L}$ [from(2)]

$\frac{2L}{3.2}=n+1$ ----------- (4)

From (3) and (4)

$\frac{2L}{3.2}=\frac{L}{2}+1$

$4L=3.2(L+2)$

$4L=3.2L+6.4$

$0.8L=6.4$

$L=8cm$