Question

A horizontal string is attached at one end to an electrical vibrator of frequency 50.0 Hz, and at the other to a set of hanging masses. The string has a mass per unit length 0.115 gm^-1. When the string has a length 210.4 cm five loops are observed. What mass (in grams) is hung from the string? You adjust the frequency so that you now observe 4 loops on the string leaving the hanging mass and string length the same. What is the new frequency?

Answer 1

Answer:

The mass of the string and new frequency are 20.78 g and 40.0 Hz.

Explanation:

Given that,

Frequency = 50.0 Hz

Mass per unit length = 0.115 g/m

String length = 210.4 cm

Number of loops = 5

We need to calculate the mass is hung from the string

Using formula of frequency

$f=\dfrac{5}{2L}\sqrt{\dfrac{T}{M}}$

$f=\dfrac{5}{2L}\sqrt{\dfrac{mg}{M}}$

Where, f = frequency

M = mass per unit length

T = tension

Put the value into the formula

$50.0=\dfrac{5}{2\times210.4}\sqrt{\dfrac{m\times9.8}{0.115}}$

$m=\dfrac{50^2\times(\dfrac{2\times210.4\times10^{-2}}{5})^2\times0.115\times10^{-3}}{9.8}$

$m=0.02078\ kg$

$m=20.78\ g$

We need to calculate the new frequency for 4 loops

Using formula of frequency

$f=\dfrac{4}{2\times210.4\times10^{-2}}\sqrt{\dfrac{0.02078\times9.8}{0.115\times10^{-3}}}$

$f=40.0\ Hz$

Hence, The mass of the string and new frequency are 20.78 g and 40.0 Hz.