Answer
given,
Frequency of oscillator = 40 Hz
Amplitude = 3 cm
linear mass density of rope = 50 x 10⁻³ kg/m
tension = T = 5 N
calculating the speed of wave
[tex]v = \sqrt{\dfrac{T}{\mu}}[/tex]
μ is linear mas density of rope
[tex]v = \sqrt{\dfrac{5}{50\times 10^{-3}}}[/tex]
v = 10 m/s
now, calculating the wavelength of the
[tex]\lambda = \dfrac{v}{f}[/tex]
[tex]\lambda = \dfrac{10}{40}[/tex]
λ = 0.25 m
now calculating transverse acceleration
a = A ω²
ω = 2 π f
ω = 2 π x 40 = 251.32 /m
a = 0.03 x 251.32²
a = 1894.96 m/s²
