Question

the displacement x of a particle at time t is given by x = 5 sin2t where x is in meters and t is in second. A simple pendulum has the same period as the particle when the length of the pendulum is
(A)10.0m
(B) 5.0m
(C) 2.5m
(D) 2.0m​

Answer 1

Answer:

Explanation:

Part (a): The wave's amplitude, wavelength, and frequency can be determined from the equation of the wave:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

The amplitude is whatever is multiplying the sine.

A = 0.9 cm

The wavenumber k is whatever is multiplying the x:

k = 1.2 m-1

The wavelength is l =  

2p

k

= 5.2 m

The angular frequency w is whatever is multiplying the t.

w = 5.0 rad/s

f =  

w

2p

= 0.80 Hz

Part (b): The wave speed can be found from the frequency and wavelength:

v = f l = 0.80 * 5.2 = 4.17 m/s

Part (c): With m = 0.012 kg/m and the wave speed given by:

v = (  

T

m

) ½

This gives a tension of T = m v2 = 0.012 (4.17)2 = 0.21 N.

Part (d): To find the direction of propogation of the wave, just look at the sign between the x and t terms in the equation. In our case we have a minus sign:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

A negative sign means the wave is traveling in the +x direction.

A positive sign means the wave is traveling in the -x direction.

Part (e): To determine the maximum transverse speed of the string, remember that all parts of the string are experiencing simple harmonic motion. We showed that in SHM the maximum speed is:

vmax = Aw

In this case we have A = 0.9 cm and w = 5.0 rad/s, so:

vmax = 0.9 * 5.0 = 4.5 cm/s

Answer 2

Explanation:

here time period of the particle is T=2π/w where w=2

so T=π we know that time period of a pendulum is T=2π√L/g where L=length of the pendulum

by putting the value of g=10m/s² we get that L=2.5m