Question

**Physics II HW 1
SHM **

- 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.

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

- 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?

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

- 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?

- What is the value of g for a location where a pendulum 1.88 m long has a period of 2.20 s?

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

- Explain SHM.

- Explain Hook’s law.

Answer #1

**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**

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