The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are 1 ms-1 and 1 ms-2 respectively. The phase velocity if he wave is 1 ms-1. The waveform is the wave is 1 ms-1. The waveform is

Question

The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are 1 ms-1 and 1 ms-2 respectively. The phase velocity if he wave is 1 ms-1. The waveform is the wave is 1 ms-1. The waveform is (A)  sin (x - t)  (B)  sin (2x - t)  (C)  sin (x -2 t)  (D)  sin (x/2 - t)

Tardigrade Admin · Accepted Answer

∵ Wave equation, y=A0 sin (k x-ω t) ...(i) where, k= angular wave number =(2 π/λ) A0= amplitude ∵ v max =a ω ∴ ω=1 (∵. given that . v max =1 (m/s), a=1 (m/s2)) ∵ k=1 (given) ∴ From Eq. (i), y= sin (x-t)

Q. The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are $1 m s^{- 1}$ and $1 m s^{- 2}$ respectively. The phase velocity if he wave is $1 m s^{- 1} .$ The waveform is the wave is $1 m s^{- 1}$ . The waveform is

$sin (x - t)$

$sin (2 x - t)$

$sin (x - 2 t)$

$sin (x /2 - t)$

$sin (x - t /2)$

$∵$ Wave equation, $y = A_{0} sin (k x - ω t)$ ...(i)
where, $k =$ angular wave number $= \frac{2 π}{λ}$
$A_{0} =$ amplitude
$∵ v_{m a x} = aω$
$∴ ω = 1$
$(∵$ given that $v_{m a x} = 1 \frac{m}{s}, a = 1 \frac{m}{s ^{2}})$
$∵ k = 1$ (given)
$∴$ From Eq. (i), $y = sin (x - t)$

Q. The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are 1ms−1 and 1ms−2 respectively. The phase velocity if he wave is 1ms−1. The waveform is the wave is 1ms−1. The waveform is

sin(x−t)

sin(2x−t)

sin(x−2t)

sin(x/2−t)

sin(x−t/2)

∵ Wave equation, y=A0​sin(kx−ωt) ...(i) where, k= angular wave number =λ2π​ A0​= amplitude ∵vmax​=aω ∴ω=1 (∵ given that vmax​=1sm​,a=1s2m​) ∵k=1 (given) ∴ From Eq. (i), y=sin(x−t)