Question

# Physics II HW 1 SHM   A 100 g object connected to a spring (k= 40 N/m)...

Physics II HW 1 SHM

1. A 100 g object connected to a spring (k= 40 N/m) oscillates on a horizontal frictionless surface with an amplitude of 4.00 cm. Find the period and the total energy of the system.

1. What is the period of a pendulum with a length of 2 meters on Earth? On the Moon?

1. A 5 kg mass is attached to a spring that is hanging vertically. The spring is stretched 0.25 m from its equilibrium position. What is the spring constant?

1. Calculate the length of a pendulum on earth whose frequency of oscillation is 10 Hz.

1. On planet X64J1, the period of a 0.50 m pendulum is 1.8 s. What is the acceleration due to gravity on this planet?
1. What is the value of g for a location where a pendulum 1.88 m long has a period of 2.20 s?

1. The period of a mass on a spring is 2 s. If k=50 N/m, what is the mass?

1. Explain SHM.

1. Explain Hook’s law.

1)
given
m = 100 g = 0.1 kg
k = 40 N/m
A = 4 cm = 0.04 m
Time period, T = 2*pi*sqrt(m/k)

= 2*pi*sqrt(0.1/40)

= 0.314 s

Total energy of the system, TE = (1/2)*k*A^2

= (1/2)*40*0.04^2

= 0.0320 J

2) given, L = 2 m

Time period of the pendulum on the Earth, T = 2*pi*sqrt(L/g)
= 2*pi*sqrt(2/9.8)
= 2.84 s

Time period of the pendulum on the moon, T = 2*pi*sqrt(L/g)
= 2*pi*sqrt(2/1.63)
= 6.96 s

3)
given
m = 5 kg
x = 0.25 m
use, F_spring = F_gravity
k*x = m*g
k = m*g/x
= 5*9.8/0.25
= 196 N/m

4) f = 10 Hz

T = 1/f = 0.1 s

we know, T = 2*pi*sqrt(L/g)

T^2 = 4*pi^2(L/g)

L = g*T^2/(4*pi^2)

= 9.8*0.1^2/(4*pi^2)

= 0.00248 m

5) on planet X64J1

use, T = 2*pi*sqrt(L/g)

1.8 = 2*pi*sqrt(0.5/g)

==> g = 6.09 m/s^2

6) use, T = 2*pi*sqrt(L/g)

2.2 = 2*pi*sqrt(1.88/g)

g = 15.3 m/s^2

7) use,

T = 2*pi*sqrt(m/k)

2 = 2*pi*sqrt(m/50)

m = 5.07 kg