<span>So this question based on doppler effect
Here apparent freq = freq of sound *(speed of sound-speed of observer)/(speed of sound-speed of source)
given
freq of sound=2000
speed of observer= 20
speed of source= 30
taking speed of sound as 343 m/s
apparent freq = 2000 [343 - 20] / [343 - 30] = 2063.90 Hz</span>
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, ![$f=\frac{\omega}{2 \pi}$](https://tex.z-dn.net/?f=%24f%3D%5Cfrac%7B%5Comega%7D%7B2%20%5Cpi%7D%24)
![$=\frac{7.9 \pi}{2 \pi}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B7.9%20%5Cpi%7D%7B2%20%5Cpi%7D%24)
= 3.95 Hz
b). The period, ![$T=\frac{1}{f}$](https://tex.z-dn.net/?f=%24T%3D%5Cfrac%7B1%7D%7Bf%7D%24)
![$T=\frac{1}{3.95}](https://tex.z-dn.net/?f=%24T%3D%5Cfrac%7B1%7D%7B3.95%7D)
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
![$x_t=4.7 \sin (7.9 \pi t)$](https://tex.z-dn.net/?f=%24x_t%3D4.7%20%5Csin%20%287.9%20%5Cpi%20t%29%24)
![$2.6 = 4.7 \sin (7.9 \pi t)$](https://tex.z-dn.net/?f=%242.6%20%3D%204.7%20%5Csin%20%287.9%20%5Cpi%20t%29%24)
![$\sin (7.9 \pi t) = \frac{26}{47}$](https://tex.z-dn.net/?f=%24%5Csin%20%287.9%20%5Cpi%20t%29%20%3D%20%5Cfrac%7B26%7D%7B47%7D%24)
![$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$](https://tex.z-dn.net/?f=%247.9%20%5Cpi%20t%20%3D%20%5Csin%5E%7B-1%7D%5Cleft%28%5Cfrac%7B26%7D%7B47%7D%5Cright%29%24)
Hence, t = 0.0236 seconds.
Answer:
given,
load = 400 N
effort = 100 N
load distance = 20 cm
we know that ,
E*Ed = L*Ld
=100 N* Ed = 400N * 20 cm
=100N * Ed = 8000N/cm
= Ed =( 8000N/cm ) / 100N
= Ed = 80 cm
b. soln.
given,
load = 600 N
load distance = 2 m
effort = 300 N
effort distance = ?
we know that ,
= E *Ed = L * Ld
= 300N * Ed = 600N * 2 m
= 300N * Ed = 1200N/m
=Ed =( 1200N/m ) / 300 N
= Ed = 4 m
C. soln.
given,
load = 600 N
load distance =20 cm
effort = 200 N
effort distance = ?
M.A = ?
V.R = ?
Efficiency = ?
we know that ,
= E *Ed = L *Ld
= 200N * Ed = 600 N * 20 cm
=200 N *Ed = 12000 N/cm
=Ed = ( 12000 N/cm) / 200 N
= Ed = 60 cm
Also,
M.A = load / effort
=600 N / 200 N
= 3
V.R = Ed/ Ld
= 60 cm / 20 cm
= 4
efficiency = ( M.A / V.R ) 100 %
= ( 3 / 4 ) 100%
= 75 %
d. soln.
given,
load = 600 N
load distance = 0.5 m
effort distance = 2.5 m
effort = ?
we know that ,
= E * Ed = L * Ld
= E * 2.5 m = 600 N * 0.5 m
= E * 2.5 m = 300 N / m
= E = ( 300 N / m ) / 2.5 m
= E = 120 N
Explanation:
The de broglie wavelength is given by :
![\lambda=\dfrac{h}{p}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7Bp%7D)
Here,
h is Planck's constant
p is momentum
Momentum and De-Broglie wavelength has inverse relationship. If momentum of an electron double, its wavelength gets half.
Answer:
6.7 m/s
Explanation:
In the vertical direction:
y₀ = 0.63 m
y = 0 m
v₀ᵧ = 0 m/s
aᵧ = -9.81 m/s²
In the horizontal direction:
x₀ = 0 m
x = 2.4 m
aₓ = 0 m/s²
Find: v₀ₓ
First, find the time:
y = y₀ + v₀ᵧ t + ½ aᵧt²
0 = 0.63 + (0) t + ½ (-9.81) t²
t = 0.358
Now, find the velocity:
x = x₀ + v₀ₓ t + ½ aₓt²
2.4 = 0 + v₀ₓ (0.358) + ½ (0) (0.358)²
v₀ₓ = 6.70
Rounded to two significant figures, the cat's velocity when it slides off the table is 6.7 m/s.