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Question

Physics Question on Wave characteristics

A wave along a string has the following equation y=0.05sin(28t1.78x)my = 0.05 \sin (28t - 1.78x)\, m (where t is in seconds and x is in meters). What are the amplitude, (A) frequency (f) and wavelength (λ)(\lambda) of the wave?

A

amplitude = 0.05 m, frequency = 4.456 Hz and wavelength λ\lambda =3.518 m

B

amplitude = 0.05 m, frequency = 28 Hz and wavelengthλ\lambda=2.0 m

C

amplitude = 5.0 m, frequency = 4.456 Hz and wavelength λ\lambda =3.518 m

D

amplitude = 0.05 m, frequency = 2.0 Hz and wavelength λ\lambda= 28 m

Answer

amplitude = 0.05 m, frequency = 4.456 Hz and wavelength λ\lambda =3.518 m

Explanation

Solution

Given the equation of wave along string
y=0.05sin(28t1.78x)y=0.05 \sin (28 t-1.78 x)...(i)
The general equation of a wave is given as
y=Asin(ωtkx)y=A \sin (\omega t-k x)...(ii)
Now, comparing Eqs. (i) and (ii), we get
Amplitude, A=0.05m,k=1.78 A=0.05\, m ,\, k=1.78
and ω=28\omega=28
\Rightarrow Frequency, f=ω2π=282π=4.456Hzf=\frac{\omega}{2 \pi}=\frac{28}{2 \pi}=4.456\, Hz
and wavelength, λ=2πk=2π1.78=3.529m\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{1.78}=3.529\, m