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The linear mass density mu of the string...

The linear mass density `mu` of the string is (where, mass of the string = m, length, of the string g = L)

A

`m/L`

B

mL

C

`mL^2`

D

`m/L_3`

Text Solution

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A
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The linear mass density of the string shown in the figure is mu = 1 g//m . One end (A) of the string is tied to a prong of a tuning fork and the other end carries a block of mass M. The length of the string between the tuning fork and the pulley is L = 2.0 m . When the tuning fork vibrates, the string resonates with it when mass M is either 16 kg or 25 kg. However, standing waves are not observed for any other value of M lying between 16 kg and 25 kg. Assume that end A of the string is practically at rest and calculate the frequency of the fork.

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Knowledge Check

  • The linear mass density 'mu' of the string is (where, mass of the string = m, length of the string = L)

    A
    `m/L`
    B
    mL
    C
    `mt^2`
    D
    `m/L_3`
  • One end of a string of length l is tied to the ceiling of a lift accelerating upwards with an acceleration g/2. The linear mass density of the string is mu(x)=mu_0x^(1//2) where, x is measured from the bottom. The time taken by a pulse to reach from bottom to top is

    A
    `sqrt((3l)/g)`
    B
    `2sqrt(l/g)`
    C
    `sqrt(l/g)`
    D
    `sqrt(l/(3g))`
  • The mass suspended from the stretched string of a sonometer is 4 kg and the linear mass density of string 4 xx 10^(-3) kg m^(-1) . If the length of the vibrating string is 100 cm , arrange the following steps in a sequential order to find the frequency of the tuning fork used for the experiment . (A) The fundamental frequency of the vibratinng string is , n = (1)/(2l) sqrt((T)/(m)) . (B) Get the value of length of the string (l) , and linear mass density (m) of the string from the data in the problem . (C) Calculate the tension in the string using , T = mg . (D) Substitute the appropriate values in n = (1)/(2l) sqrt((T)/(m)) and find the value of 'n' .

    A
    BCAD
    B
    ABCD
    C
    DCBA
    D
    BADC
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