An oscillating block-spring system has a mechanical energy of 1.00 J, an amplitu

Question

An oscillating block-spring system has a mechanical energy of 1.00 J, an amplitude of 11.7 cm, and a maximum speed of 1.16 m/s. Find the spring constant. Find the mass of the block. Find the frequency of oscillation. If the phase angle for a block-spring system in SHM is pi/7 rad and the block's position is given x = x_m cos (omega t + phi), what is the ratio of the kinetic energy to the potential energy at time t = 0? If y(x, t) = (4.8 mm) sin(kx + (595 rad/s)t + phi) describes a wave traveling along a string, how much time does any given point on the string take to move between displacements y = -2.0 mm and y = +2.0 mm?

Explanation / Answer

IN SHM

total energy E = (1/2)*K*A^2

1 = (1/2)*K*0.117^2

k = 146.1 N/m <<<<<---------------answer

(b)

total energy E = max KE

E = (1/2)*m*Vmax^2

1 = (1/2)*m*1.16^2

m = 1.5 kg

c)

In SHM

x = A*cos(wt)

speed v = dx/dt = A*sinwt*w

V = A*w*sinwt

v = A*w*sqrt(1-cos(wt)^2)

v = A*w*sqrt(1-x^2/A^2)

v = w*sqrt(A^2-x^2)

v is maximum when x = 0

vmax = A*w

1.16 = 0.117*2*pi*f

f = 1.6 Hz <<<----------answer

++++++++++++++++++++++

v = dx/dt = -xm*w*sin(wt+phi)

at t = 0

vo = -Xm*w*sinphi

kinetic energy K = (1/2)*m*w^2*xm^2*(sin(phi))^2.............(1)

potential energy

at t= 0

x = Xm*cosphi

U = 0.5*k*x^2

Uo = 0.5*k*Xm^2*(cosphi)^2

k = m*w^2

Uo = 0.5*m*w^2*Xm^2*(cosphi)^2..........2)

1/2

Ko/Uo = (tanphi)^2 = 0.232 <<<<<----------------answer

+++++++++++++++++++++++++++++++++

for particle at x = 0 at time t1 = 0 its displacement

y1 = -2 mm

-2 = 4.8*sin(phi)

at t2 = t

at y = + 2mm

2 = 4.8*sin(595t+phi)

sin(595t+phi) = -sin(phi)

sin(595t+phi) = sin( pi + phi)

595t + phi = pi + phi

595t = pi

t = pi/595

t = 0.0053 s <<<<-------------answer

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