The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book
height(h)=1 m
time taken=0.64 s


![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley


And Tension in Rope


T=8.364 N
and Tension will provide Torque




Thus mass is uniformly distributed or some more towards periphery of Pulley
Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
= Coefficient of friction = 0.4
Energy stored in spring is given by

As the energy in the system is conserved we have

The speed of the 8 kg block just before collision is 3.258 m/s
Answer:
speed = distance/time
Explanation:
speed = 150/30
speed =5m/s
you were running fast .....5m/s is a good speed