Answer:
a)
Nm⁻¹
b)
Hz
c)
ms⁻¹
d)
m
e)
ms⁻²
f)
m
Explanation:
a)
= force required to hold the object at rest connected with stretched spring = 27 N
= stretch in the spring from equilibrium position = 0.2 m
= force constant of the spring
force required to hold the object at rest is same as the spring force , hence
![F = k x](https://tex.z-dn.net/?f=F%20%3D%20k%20x)
![k = \frac{F}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7BF%7D%7Bx%7D)
inserting the values
Nm⁻¹
b)
frequency of the oscillations is given as
![f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=f%20%3D%5Cfrac%7B1%7D%7B2%5Cpi%20%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
inserting the values
Hz
c)
= Amplitude of oscillations = 0.2 m
= angular frequency
Angular frequency is given as
rads⁻¹
Maximum speed of oscillation is given as
![v_{max} = Aw](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20Aw)
ms⁻¹
d)
maximum speed of the object occurs at the equilibrium position, hence
m
e)
Maximum acceleration of oscillation is given as
![a_{max} = Aw^{2}](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20Aw%5E%7B2%7D)
ms⁻²
f)
maximum acceleration occurs when the object is at extreme positions, hence
m
Answer:
Tension, T = 2038.09 N
Explanation:
Given that,
Frequency of the lowest note on a grand piano, f = 27.5 Hz
Length of the string, l = 2 m
Mass of the string, m = 440 g = 0.44 kg
Length of the vibrating section of the string is, L = 1.75 m
The frequency of the vibrating string in terms of tension is given by :
![f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7B1%7D%7B2L%7D%5Csqrt%7B%5Cdfrac%7BT%7D%7B%5Cmu%7D%7D)
![\mu=\dfrac{m}{l}](https://tex.z-dn.net/?f=%5Cmu%3D%5Cdfrac%7Bm%7D%7Bl%7D)
![\mu=\dfrac{0.44}{2}=0.22\ kg/m](https://tex.z-dn.net/?f=%5Cmu%3D%5Cdfrac%7B0.44%7D%7B2%7D%3D0.22%5C%20kg%2Fm)
![T=4L^2f\mu](https://tex.z-dn.net/?f=T%3D4L%5E2f%5Cmu)
![T=4\times (1.75)^2\times (27.5)^2 \times 0.22](https://tex.z-dn.net/?f=T%3D4%5Ctimes%20%281.75%29%5E2%5Ctimes%20%2827.5%29%5E2%20%5Ctimes%200.22)
T = 2038.09 N
So, the tension in the string is 2038.09 N. Hence, this is the required solution.
Answer:
Using Faraday's law;
ε= -N ∆ψ(B)/ ∆t;
∆t= -N ∆ψ(B)/ ε
Explanation:
Using Faraday's law; Faraday's law state that the induced emf is directly proportional to the rate of change of time of magnetic flux
ε= -N ∆ψ(B)/ ∆t;
Where ε= induced EMF, ∆ψ(B)/ ∆t is the rate of change of magnetic flux, ψ(B) = BA cos θ
θ= the angle between the magnetic field B and the normal surface area.
We can also calculate the direction of induced magnetic flux. At first, the field is perpendicular to the plane of the loop,the loop can rotate about either an horizontal or vertical axis passing through the mid point