In the arrangement in the figure, a string, tied to a sinusoidal oscillator at P

Question

In the arrangement in the figure, a string, tied to a sinusoidal oscillator at P and running over a support at Q, is stretched by a block of mass m. The separationLbetween P and Q is 1.22 m, and the frequency f of the oscillator is fixed at 120 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging block is 272 g or 425 g, but not for any intermediate mass. What is the linear density ? of the string?

Explanation / Answer

k is linear density

m1 = 272 g = 0.272 kg

m2 = 425 g = 0.425 kg

f = 120 Hz

f = (n+1) vn / 2 L where vn = sqrt(m1 g / k )

f = n vn+1 / 2L where vn = sqrt(m2 g / k )

2 L f / n+1 = sqrt( 0.272 g / k )

2 L f / n = sqrt(0.425 g / k )

n /(n+1) = sqrt(0.272/0.425)

1 + 1/n = sqrt(0.425/0.272)

1/n =  sqrt(0.425/0.272) - 1

2 L f / n = sqrt(0.425 g / k )

2 *1.22*120*(sqrt(0.425/0.272) - 1) = sqrt(0.425*9.81 / k )

Solving above equation we get , k = 0.000778 kg/m = 0.778 g/m

---------------------------------------------------------------------------------------------------------------------------------------------

Linear density of string = 0.778 g / m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.