Answer to Question #195537 in Mechanics | Relativity for Mohammad Balhas

Question

The left end of a taut string of length L, is connected to a vibrator with a fixed frequency f. The right end of the string is tied to a suspended object of varying mass, m, through a pulley. For a mass m1 of the object, a standing wave with one loop is observed. For a mass m2 of the object, a standing wave with two loops is observed. What is the ratio "m2 / m1 = ?"

0.0625

0.25

2

4

16

Answer 1

Gives string length =L

Frequency=n

Mass (m)=m₁

Loop(n)=1

Mass(m)=m₂

Loop(n)=2

Case (1)

$L=\frac{n\lambda}{2}$

n=1

$L=\frac{\lambda}{2}$

$\lambda=2L$

$v=\sqrt\frac{T}{m_1}$

$\frac{T}{m_1}=({n\lambda})^2$

$\frac{T}{m_1}=({n\times 2L})^2$

$m_1=\frac{T}{v^2}$

$m_1=\frac{T}{4(n\lambda)^2}$

Case(2)

n=2

$L=\frac{2\lambda}{2}$

$L=\lambda$

$\frac{T}{m_2}=({n\times L})^2$

$m_2=\frac{T}{(n\lambda)^2}$

$\frac{m_2}{m_1}=\frac{\frac{T}{(n\lambda)^2}}{\frac{T}{4(n\lambda)^2}}=\frac{4}{1}$