The equation of a transverse wave travelling on a rope is given by y-10sin- 0-01x-2- t where y and x are in cm and t in seconds- The maximum transverse speed of a particle in the rope is about-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Different Forms of Equations

The equation ...

Question

The equation of a transverse wave travelling on a rope is given by $y = 10 sin π (0.01 x - 2 π t)$ where $y$ and $x$ are in $c m$ and $t$ in seconds. The maximum transverse speed of a particle in the rope is about.

63 c m / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

75 c m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 c m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

121 c m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $63 c m / s$
The given equation is $y = 10 sin (0.01 π x - 2 π t)$
Hence, $ω =$ coefficient of $t = 2 π$
$\Rightarrow$ Maximum speed of the particle, $v_{m a x} = a ω$
$= 10 \times 2 π$
$= 10 \times 2 \times 3.14$
$= 62.8 \approx 63 c m / s$

Suggest Corrections

Similar questions

The equation of a transverse wave travelling on a rope is given by $y = 10 s i n π (0.001 x - 2.00 t)$ where y and x are in centimeters and t in seconds. The maximum transverse speed of a particle in the rope is about.

Q. The equation of a transverse wave is given by

y = 10 s i n π (0.01 x - 2 t)

where x and y are in cm and t is in second. its frequency is

Q. The equation of the transverse wave travelling in a rope is given by

y = 5 sin (4 t - 0.02 x)

where

y

and

x

are in meters and

t

is in seconds. Calculate the intensity of the wave if the density of rope material is

1250 k g / m^{3}

The equation of a transverse wave is given by $y = 10 s i n π (0.01 x - 2 t)$ where x and y are in cm and t is in second. Its frequency is

Q. At t = 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a wire is described by the function

y = \frac{6}{x^{2}} g i v e n t h a t x \neq 0

. Transverse velocity of a particle at x = 2 m and t = 2 s is

Join BYJU'S Learning Program

The equation of a transverse wave travelling on a rope is given by y=10sinπ(0.01x−2πt) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about.

The correct option is B 63 cm/sThe given equation is y=10sin(0.01πx−2πt)Hence, ω=coefficient of t=2π⇒ Maximum speed of the particle, vmax=aω=10×2π=10×2×3.14=62.8≈63cm/s

The equation of a transverse wave travelling on a rope is given by $y = 10 sin π (0.01 x - 2 π t)$ where $y$ and $x$ are in $c m$ and $t$ in seconds. The maximum transverse speed of a particle in the rope is about.

The correct option is B $63 c m / s$
The given equation is $y = 10 sin (0.01 π x - 2 π t)$
Hence, $ω =$ coefficient of $t = 2 π$
$\Rightarrow$ Maximum speed of the particle, $v_{m a x} = a ω$
$= 10 \times 2 π$
$= 10 \times 2 \times 3.14$
$= 62.8 \approx 63 c m / s$