Question

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string’s mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

Answer 1

Answer:

The string's mass and the maximum transverse acceleration are 2.2 g and 22400.51 m/s².

Explanation:

Given that,

Length of string = 2.50 m

Tension = 90.0 N

Amplitude = 3.50 cm

Speed = 28.0

First overtone ,

$\lambda =l$

(a). We need to calculate the mass of string

Using maximum transverse speed at antinodes

$v_{max}=A\omega$

$A\omega=28$

$A\times2\pi f=28$

Put the value into the formula

$f=\dfrac{28}{2\times3.14\times3.50\times10^{-2}}$

$f=127.39\ Hz$

Using formula of wavelength

$v = f\lambda$

$\sqrt{\dfrac{T}{\mu}}=f\lambda$

$\mu=\dfrac{90}{(127.39\times2.50)^2}$

$\mu=8.8734\times10^{-4}$

Mass of string = $\mu\times l$

Mass of string = $8.8\times10^{-4}\times2.50$

Mass of string =2.2 g

(b). We need to calculate the maximum transverse acceleration of this point on the string

Using formula of the maximum transverse acceleration

$a=A\omega^2$

$a=A\times(2\pif)^2$

Put the value into the formula

$a=3.50\times10^{-2}(\times2\times3.14\times127.39)^2$

$a=22400.51\ m/s^2$

Hence, The string's mass and the maximum transverse acceleration are 2.2 g and 22400.51 m/s².